Analysis II

Last updated Nov 1, 2014

Analysis, Volume II
Terence Tao
Hindustan Book Agency, January 2006. Third edition, 2014
Hardcover, 236 pages.

ISBN 81-85931-62-3 (first edition)

This is basically an expanded and cleaned up version of my lecture notes for Math 131B. In the US, it is available through the American Mathematical Society. It is part of a two-volume series; here is my page for Volume I.

— Errata to the first edition (softcover) —

p. 392, example 12.1.7: $5+2=7$ should be $3+4=7$ .
p. 393, example 12.1.9: $\sup(5,2)=7$ should be $\sup(3,4)=4$ .
p. 394, example 12.1.13: (iii) should be (c).
p. 403, example 12.2.13: delete the redundant “, but not the other”.
p. 404, line 4: “neither open and closed” should be “neither open nor closed”.
p. 415, line 3: $\alpha \in A$ should be $\alpha \in I$ .
p. 416, line 11: “ $k \geq j$ ” should be “ $k > j$ “.
p. 419, line -2: In Exercise 12.5.15, = should be $\neq$ . Also, “that by counterexample” should be “by counterexample that”
p. 426, Exercise 13.2.9: $X$ should be ${\Bbb R}$ throughout. Also, the definition of limsup and liminf for functions has not been given; it can be reviewed here, e.g. by inserting “where we define $\limsup_{x \to x_0} f(x) := \inf_{r>0} \sup_{x: |x-x_0| \leq r} f(x)$ and $\liminf_{x \to x_0} f(x) := \sup_{r>0} \inf_{x: |x-x_0| \leq r} f(x)$ .”
p. 435, Definition 13.5.6: “metric space” should be “topological space”.
p. 438, Exercise 13.5.9: One needs to assume as an additional hypothesis that X is first countable, which means that for every x in X there exists a countable sequence V_n of neighborhoods of x, such that every neighbourhood of x contains one of the V_n.
p. 452, Exercise 14.3.6: “Propositoin” should be “Proposition”.
p. 452, Exercise 14.3.8: “ $x \in {\Bbb R}$ ” should be “ $x \in X$ “.
p. 458: Exercise 14.5.2 should be deleted and redirected to Exercise 14.2.2(c).
p. 459: In line 11, $2 \epsilon$ should be $\epsilon$ .
p. 464: ) missing at the end of Exercise 14.7.2. An additional exercise, Exercise 14.7.3 is missing; it should state “Prove Corollary 14.7.3.”.
p. 466: Exercise 14.8.8 should be Exercise 14.8.2.
p. 467: Exercise 14.8.11 should be Exercise 14.8.4.
p. 469: “Limits of integration” should be “Limits of summation”. In Lemma 14.8.14, $3M+2\delta$ should be $1+4M$ , and Exercise 14.8.14 should be Exercise 14.8.6.
p. 470: Exercise 14.8.15 should be Exercise 14.8.7. Exercise 14.8.16 should refer to a (currently non-existent) Exercise 14.8.9, which of course would be to prove Lemma 14.8.16.
p. 471: At the end of the proof of Corollary 14.8.19, $|P(y)-g(y)| \leq \varepsilon$ should be $|P(y)-f(y)| \leq \varepsilon$ .
p. 472: In Exercise 14.8.2(c), Lemma 14.8.2 should be Lemma 14.8.8.
p. 477: In Exercise 15.1.1(e), Corollary 14.8.18 should be Corollary 14.6.2.
p. 478: In Example 15.2.2, $\frac{1}{x-1}$ should be $\frac{1}{1-x}$ .
p. 482: In Exercise 15.2.5, the $(x-b)^n$ on the right-hand side should be $(x-b)^m$ .
p. 486: In second and third display, y should be in ${}[0,1]$ rather than $(-1,1)$ .
p. 493: In Exercise 15.5.4, $x < 0$ should be $x \leq 0$ .
p. 501: In Theorem 17.7.2, “if $f(x_0)$ is not invertible” should be “if $f'(x_0)$ is not invertible”.
p. 502: In Exercise 15.6.6, Lemma 15.6.6 should be Lemma 15.6.11.
p. 511: “Fourier… was, among other things, the governor of Egypt during the reign of Napoleon. After the Napoleonic wars, he returned to mathematics.” should be “Fourier… was, among other things, an administrator accompanying Napoleon on his invasion of Egypt, and then a Prefect in France during Napoleon’s reign.”
p. 556: In Theorem 17.5.4, f can take values in ${\Bbb R}^m$ and not just in $R$ ; insert the line “By working with one component of $f$ at a time, we may assume $m=1$ ” as the first line of the proof. Also, $f(x-x_0)$ should be $f(x+x_0)$ .
p. 557: In the second display, $-f(0)$ should be $+f(0)$ .
p. 560: In Exercise 17.6.1, add the hypothesis “and $f'$ is continuous” before “, show that $f$ is a strict contraction”.
p. 561: In Exercise 17.6.3, change “which is a strict contraction” to “such that $|f(x)-f(y)| < |x-y|$ for all distinct $x,y$ in ${}[a,b]$ “. In Exercise 17.6.8, $\max(c,c')$ should be $\min(c,c')$ .
p. 562: In Theorem 17.7.2, $T: E \to {\Bbb R}^n$ should be $f: E \to {\Bbb R}^n$ .
p. 565, line -7: $U$ should be $f^{-1}(B(0,r/2))$ rather than $f^{-1}(B(0,r))$ .
p. 570, first display: all partial derivatives should have a – sign (not just the first one). Last paragraph: “Thus $(x_1,\ldots,x_{n-1})$ lies in W” should be “Thus $(x_1,\ldots,x_{n-1})$ lies in U”.
p. 571, second display: add “ $=0$ ” at the end.
p. 584, Corollary 18.2.7: “ $x_i \in [a_i,b_i]$ ” should be “ $x_i \in (a_i,b_i)$ “.
p. 599, Definition 18.5.9: $(a,\infty)$ should be $(a,+\infty]$ .
p. 600: In Lemma 18.5.10, $f: \Omega \to {\Bbb R}$ should be $f: \Omega \to {\Bbb R}^*$ . In the second and fourth lines of the proof of this lemma, $(a,+\infty)$ should be $(a,+\infty]$ .
p. 616-617, Exercise 19.2.10: ${\Bbb R}$ should be ${}[0,1]$ throughout.

— Errata to the second edition (hardcover) —

p. 351, At the end of Example 12.1.6, add “Extending the convention from Example 12.1.4, if we refer to ${\bf R}^n$ as a metric space, we assume that the metric is given by the Euclidean metric unless otherwise specified.”
p. 372, In Case 1 of the proof of Theorem 12.5.8, all occurrences of “ $y^{(n)}$ should be $y^{(n_j)}$ in the second paragraph.
p. 374, In Exercise 12.5.12(b), the phrase “with the Euclidean metric” should be deleted.
p. 390: In Exercise 13.5.5, “there exist $a, b \in X$ such that the “interval” $\{ y \in X: a < y < b \}$ ” should be replaced with “there exists a set $I$ which is an interval $\{y \in X: a < y < b\}$ for some $a, b \in X$ , a ray $\{y \in X: a < y \}$ for some $a \in X$ , the ray $\{ y \in X: y < b\}$ for some $b \in X$ , or the whole space $X$ , which”. In Exercises 13.5.6 and 13.5.7, “Hausdorff” should be “not Hausdorff”.
p. 390: Exercise 13.5.8 should be replaced as follows: “Show that there exists an uncountable well-ordered set $\omega_1+1$ that has a maximal element $\infty$ , and such that the initial segments $\{ x \in \omega_1+1: x < y \}$ are countable for all $y \in \omega_1+1 \backslash \{\infty\}$ . (Hint: Well-order the real numbers using Exercise 8.5.19, take the union of all the countable initial segments, and then adjoin a maximal element $\infty$ .) If we give $\omega_1+1$ the order topology (Exercise 13.5.5), show that $\omega_1+1$ is compact; however, show that not every sequence has a convergent subsequence.”
p. 395: In Proposition 14.1.5(d), add “Furthermore, if $x_0 \in E$ , then $f(x_0)=L$ .”
p. 396: In Exercise 14.1.5, $\lim_{y \to y_0; y \in f(E)} g(x)$ should be $\lim_{y \to y_0; y \in f(E)} g(y)$ , and $\lim_{x \to x_0; x \in E} g \circ f(x_0)$ should be $\lim_{x \to x_0; x \in E} g \circ f(x)$ .
p. 425: In Theorem 15.1.6(d), the summation should start from n=1 rather than n=0.
p. 427: Just before Definition 15.2.4, “for some $a \in {\ bf R}$ ” should be “for some $r > 0$ “.
p. 431: In Exercise 15.2.8(e), “ $d_m (x-b)^n$ ” should be “ $d_m (x-b)^m$ . In Exercise 15.2.8(d), $(s-\varepsilon)^m$ should be $(s-\varepsilon)^{-m}$ .
p. 433 (proof of Theorem 15.3.1): $s_N$ in the third display and $s_0$ in the next line should be
$S_N$ and $S_0$ respectively.
p. 452: In Exercise 15.7.2, $x-y$ should be $x_0-y$ . In Exercise 15.7.6, “complex real number” should be “complex number”.
p. 473: In Exercise 16.5.4, $in \hat f(n)$ should be $2\pi i n \hat f(n)$ .
p. 477: In Example 17.1.7, $x \in {\Bbb R}$ should be $c \in {\Bbb R}$ .
p. 486: In Definition 17.3.7, $1 \leq j \leq m$ should be $1 \leq j \leq n$ , and $x_0+tv$ should be $x_0+te_j$ .
p. 488: In the definition of L in the proof of Theorem 17.3.8, m should be n.
p. 492: In Exercise 17.3.1, Exercise 17.1.3 should be Exercise 17.2.1.
p. 495: In the proof of Theorem 17.5.4, $a'$ should equal $\frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} f(0)$ rather than $\frac{\partial}{\partial x_j} \frac{\partial}{\partial x_i} f(0)$ .
p. 499, proof of Lemma 17.6.6: After “ $F$ does indeed map $B(0,r)$ to itself.”, add “The same argument shows that for a sufficiently small $\varepsilon > 0$ , $F$ maps the closed ball $\overline{B(0,r-\varepsilon)}$ to itself. After “ $F$ is a strict contraction”, add “on $B(0,r)$ , and hence on the complete space $\overline{B(0,r-\varepsilon)}$ “.
p.502, proof of Theorem 17.7.2: “ $f^{-1}(y)={\tilde f}^{-1}(y+f(x_0))$ ” should be “ $f^{-1}(y)={\tilde f}^{-1}(y-f(x_0))$ “.
p. 505, Section 17.8: $(0,1)$ should be $(-1,0)$ . In the second paragraph, the function $f$ should be $g$ (for better compatibility with the discussion of the implicit function theorem).
p.508, proof of Theorem 17.8.1, “U is open and contains $(y_1, \dots,y_{n-1},0)$ ” should be “U is open and contains $(y_1, \dots,y_{n-1})$ “.
p. 515: In the display before Definition 18.2.4, $j=\in J$ should be $j \in J$ . In Definition 18.2.4, $\sum_{j=1}^\infty$ should be $\sum_{j \in J}$ .
p. 520: In Example 18.2.9, $\sum_{q\in\mathbf{Q}} m*(\mathbf{Q})$ should be $\sum_{q\in\mathbf{Q}} m*(\{q\})$ in the display.
p. 528, proof of Lemma 18.4.8: On the second line, “let $A$ be any other measurable set” should be “let $A$ be an arbitrary set (not necessarily measurable)”.
p. 545: In Corollary 19.2.11, “non-negative functions” should be “non-negative measurable functions”.
p. 555, Remark 19.5.2: x and y should be swapped in “equals 1 when $x>0$ and y=0, equals -1 when $x<0$ and y=0, and equals zero otherwise”.

– Errata for the third edition (hardcover) –

Page 42: In Exercise 2.5.5, $V \subset X$ should be $V \subseteq X$ . In Exercise 2.5.13, “topological space” should be “Hausdorff topological space”.
Page 98, 100: Exercises 4.6.10-4.6.13 may be deleted, and the paragraph after Lemma 4.6.13 may be replaced with “Observe that with our choice of definitions, the space ${\bf C}$ of complex numbers is identical (as a metric space) to the Euclidean plane ${\bf R}^2$ , since the complex distance between two complex numbers $(a,b), (a',b')$ is exactly the same as the Euclidean distance $\sqrt{(a-a')^2+(b-b')^2}$ between these points. Thus, every metric property that ${\bf R}^2$ satisfies is also obeyed by ${\bf C}$ ; for instance, ${\bf C}$ is complete and connected, but not compact.”
Page 122: In Remark 5.5.2, “continuously differentiable” may be relaxed to just “differentiable”, and “twice continuously differentiable” may be relaxed to “continuously differentiable”.
Page 176: In Proposition 7.3.3, $\sum_{q \in J} q+E$ should be $\bigcup_{q \in J} q+E$ (two occurrences).
Page 178: In Corollary 7.4.7, $m(B \backslash A) = m(B) - m(A)$ should be $m(B \backslash A) + m(A) = m(B)$ .

Caution: the page numbering is not consistent across editions. In the third edition, the chapters were renumbered to start from 1, rather than from 12.

Thanks to Biswaranjan Behera, José Antonio Lara Benítez, Carlos, EO, Florian, Gökhan Güçlü, Bart Kleijngeld, Eric Koelink, Wang Kunyang, Matthis Lehmkühler, Jason M., Manoranjan Majji, Geoff Mess, Cristina Pereyra, Kent Van Vels, Haokun Xu, and the students of Math 401/501 and Math 402/502 at the University of New Mexico for corrections.

180 comments

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23 November, 2007 at 10:37 am

Anonymous

Dear Terry,

The AMS url in your post links to another book rather than “Analysis II.” Additionally, I was trying to order the book from Amazon, and was only able to order “Analysis I” through Amazon directly, but couldn’t do the same for “Analysis II”, even though a page exists for the second volume. Any plans on it being sold *directly* by Amazon?

Thanks!

23 November, 2007 at 12:02 pm

Terence Tao

Dear Anonymous,

Thanks for the bad link report. I just checked Amazon and the second volume is currently available (well, there are two copies available, at least):

Analysis II (Texts and Readings in Mathematics, No. 38) (v. 2)

But it is a rather low volume item, so availability is likely to continue to be spotty.

20 February, 2009 at 4:30 am

carlos

Dear Professor Tao,

First of all I would like you to congratulate you for the two volumes of the book. I had been looking for a reference to revise calculus from scratch and they have proven invaluable.

I would like to point out a few other minor errata I noticed:

p. 471 at the end of the proof |P(y)-g(y)| should be |P(y)-f(y)|
p.477 ex. 15.1.1 corollary 14.6.2 seems a better hint for (e)
p.482 ex.15.2.5 at the end of the identity, the last exponent should be m and not n

I also have a small question concerning Ex. 15.1.1 in page 477. To prove (c) the hint is to use the Weierstrass M-test. However, in the case of a function with an infinite convergence radius (such as for example sin(x) ), the sup norm of each of the f_n is infinite and consequently their infinite sum is not convergent, so I don’t see how the test can be applied. I am probably just making some basic mistake and I would really appreciate it if someone could point it out…

20 February, 2009 at 9:09 am

Terence Tao

Dear Carlos,

Thanks for the corrections! For Ex. 15.1.1(c), one is only asking for uniform convergence on a compact interval [a-r,a+r] (r can be arbitrarily large, but is finite), and in this case the summands are absolultely convergent.

22 March, 2009 at 10:37 am

Florian

Dear Professor Tao,

firstly, I can only agree with Carlos and I congratulate you to your book as well. I especially like the structure of it because you emphasize the importance of the basic elements of analysis (which aren’t obvious at all…).

I have a question concerning your example on page 625. Is the given example function f(x,y) really not (absolutley) integrable over y in x=0?
It would make sense to me, if it wasn’t integrable over x in y=0. Or am I wrong?

Best regards,
Florian

24 March, 2009 at 2:05 pm

Terence Tao

Oops, you’re right, it should be switched. Thanks!

22 April, 2009 at 10:17 pm

Mark

Professor Tau,
I am attempting exercise 12.5.14 in Analysis II and would like to know if you could elaborate on the hint given.

Any help would be greatly appreciated.

Thank you.
Mark

7 March, 2013 at 10:00 pm

Luqing Ye

Dear Mark,

I provide my hint in 2013,though your question is in 2009…my hint is essentially same to Mr.Tao’s hint,but I express it differently.

Because for all the point $x$ in E , $d(x_0,x)$ is non-negative,so the set of all $d(x_0,x)$ has a greatest lower bound $l$ which is non-negative.If this greatest lower bound can be reached by a point $p$ in E,such that $d(p,x_0)=l$ ,then done.Otherwise,【there is a sequence of elements $m_1,m_2,\cdots,m_n,\cdots$ in E,such that $\lim_{n\to\infty}d(m_n,x_0)=l$ 】.Because E is compact,so this sequence has a convergent subsequent,suppose that this subsequence convergent to $k$ in E,now we prove that $d(k,x_0)=l$ .If not,then $d(k,x_0)>l$ ,then this contradict to the words in 【】.

23 April, 2009 at 12:16 pm

Terence Tao

Dear Mark,

By compactness, the sequence $x^{(n)}$ in the hint has a subsequence which converges to a limit $x$ in E. Now compute what the distance of x to $x_0$ is.

7 March, 2013 at 10:25 pm

Luqing Ye

Dear Prof.Tao,

The only difference between my hint and your hint is that you introduce $d(x_0,x^(n))\leq R+\frac{1}{n}$ ,but I make some skip,I directly say $\lim_{n\to\infty}d(m_0,x_0)=l$ .I think this skip make my argument requires the axiom of choice….

No……I have to rethink,I think that your argument also require the axiom of choice…we need the axiom of choice to construct that sequence.

Is that true?

8 March, 2013 at 9:54 pm

Luqing Ye

Maybe I need to answer my question myself.I hope my comments do not disturb you.

Actually,lemma 8.4.5 has already answered this question.When E is a closed set,we don’t need the axiom of choice.Prof.Tao’s argument do not require the axiom of choice,my argument is nothing,because I really make some skip!I need to explain why the words in 【】is true.

23 April, 2009 at 5:51 pm

Mark

Dear Professor Tau,

Thank you for taking the time to assist me.
Your help has been greatly appreciated.

Mark

30 April, 2009 at 9:15 am

Jorge

Dear Professor Tao,

I’m trying to crack Ex. 14.8.2. in your Analysis II but I’m stuck on the part c), namely on the estimation|c*(1 – x^2)^N| <= epsilon which must hold for delta <= |x| <= 1 (this is precisely the condition from part c) in the definition of the approximation to the identity, I mean…). How can I now use part b) of the same Ex. here? Or this is not the case?

Thank you.
Jorge

30 April, 2009 at 10:06 am

Terence Tao

Dear Jorge,

Part (b) ensures that the normalising constant c only grows polynomially in N (or more precisely, is bounded by $C \sqrt{N}$ for some C). On the other hand, once x is bounded away from zero, $(1-x^2)^N$ decays exponentially in N (uniformly in x). As exponential decay always beats polynomial growth, the integral of $c (1-x^2)^N$ away from zero will go to zero.

30 April, 2009 at 10:36 pm

Jorge

So that f_N(x) := c(1-x^2)^N goes to zero function uniformly as N tends to infinity?

1 May, 2009 at 8:54 am

Terence Tao

On any set of the form $\{x: \varepsilon \leq |x| \leq 1 \}$ , yes.

1 May, 2009 at 9:15 am

Jorge

Thank you very much for your time. It is really great thing to be able to consult with you online.

Jorge

23 May, 2009 at 6:31 am

HASSAN JOLANY

HELLO I AM STUDENT UNIVERSITY OF TEHRAN I STUDY ABOUT THE POINCARE CONJECTURE FOR MY THESIS CAN I ASK YOU SOME OF MY PROBLEMS
BEST REGARD FOR YOU

1 June, 2009 at 10:09 pm

Joe

Dear Professer Tao,
I am attempting exercise 14.2.1 (a) from Analysis II. What do I need to show? For the “only if” direction I think I need to show that if f is continuous and the sequence goes to zero then I have pointwise convergence (is this right?), but I am not sure about the “if” direction. Can you give me any help? Thank you,

– Joe

2 June, 2009 at 8:42 am

Mark

Dear Prof Tao,

Is there any way I could get or browse the answers for all the exercises in Analysis 2?

2 June, 2009 at 9:43 am

Terence Tao

Dear Joe,

Yes, this is the “only if” part. The “if part” asserts that if f is a function such that $f_{a_n}$ converges pointwise to f for every sequence $a_n$ converging to zero, then f is necessarily continuous.

Dear Mark: I do not have an answer key for the exercises. Some of the exercises are based on my classes at

http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/

which have a partial solution set, but this is far from complete (also, some of the wording of the exercises changed when converted to book form).

2 June, 2009 at 5:05 pm

mark

Thank you prof. Another thing is there any solutions for all the homeworks of your Math131BH?

2 June, 2009 at 7:28 pm

Ali Nesin

Hi prof,

I do come across this q and really need your advise.
Suppose (X,d1), (X,d2) are two metric spaces over the same set X, and c1, c2 are strictly positive.
How do we show that (X,d), where d = c1d1 + c2d2 is also metric space?

2 June, 2009 at 8:44 pm

timur

Dear Ali Nesin,

Try going over the metric space axioms d has to satisfy. Which axiom(s) are you having trouble with?

3 June, 2009 at 1:02 am

Ali Nesin

I do have trouble actually in understanding the q. I think it is the real line, R I’m dealing with. Am I right?

3 June, 2009 at 1:18 pm

timur

Dear Ali Nesin,

I think you should ask your question on a forum like the following (I hope prof. Tao’s does not mind):

http://www.sciforums.com/

http://www.mathlinks.ro/Forum/

You even might find your question already answered.

5 June, 2009 at 6:31 pm

Mark

Mark

Professor Tau,
I am attempting exercise 14.8.8 in Analysis II and would like to know if you could elaborate on the hint given.

Any help would be greatly appreciated.

Thank you.
Mark

6 June, 2009 at 2:12 pm

Anonymous

In the proof of Abel’s Theorem on page 487 is the limit superior suppose to be as y approaches 1 instead of as n approaches infinity?

6 June, 2009 at 8:12 pm

Mark Werner

This is Mark Werner.

I am the original ‘Mark’ in this blog who asked about question 12.5.14. I am not the most recent ‘Mark’ asking about question 14.8.8 which has been set as a take home exam question with no consultation allowed. I believe someone in the class is pretending to be me, since the question was asked in exactly the same way as my original one, including the misspelling your name. I am posting this just to ensure that my name is clear.

Thank you.

– Mark

7 June, 2009 at 10:29 am

student

Dear Prof Tao,

is it possible to find a 1-1 map from [0,1] to (0,1)?

thanks

7 June, 2009 at 10:44 am

timur

Dear student,

There is a general theorem in set theory (called Cantor-Bernstein theorem) that says that for two sets A and B if there are injective maps A -> B and B -> A then there exists a bijective map A -> B. I believe Cantor proved it using AC and later Bernstein gave a constructive proof without AC.

7 June, 2009 at 2:39 pm

student

Thank you Timur,

actually I want to apply that theorem, there is an obvious 1-1 map from (0,1) to [0,1] but I could not find a 1-1 map from [0,1] to (0,1).

if anyone knows, please let me know

thanks

7 June, 2009 at 3:07 pm

timur

Sorry, I misunderstood your question; you can scale and translate [0,1] to, say, [z,1-z] for some z>0 to get an injective map [0,1] -> (0,1).

7 June, 2009 at 3:15 pm

student

good idea thanks Dear Timur

7 June, 2009 at 7:41 pm

geometry beginner

can anyone suggest me a nice introductory book on differential geometry and algebraic topology? my background is just point set topology.

if you know nice introductory survey articles as well, please let me know.

thanks

18 July, 2009 at 5:02 pm

Jason M.

Hello — I had a couple of very small observations:

On page 546, in Definition 17.3.7 (Partial derivative), $f$ is a function defined on a subset of $R^n$, and you are defining the partial derivative with respect to $x_j$, but you index $j$ from 1 to $m$ instead of from 1 to $n$.

Later in that same definition, you write $\lim_{t\to 0;t\neq 0;x_0+tv\in E}$, when I think it ought to be $x_0+te_j$ instead of $x_0+tv$.

Thanks! –Jason

18 July, 2009 at 5:49 pm

Jason M.

Hello again — I have noticed one more thing:

On page 548, in the proof of Theorem 17.3.8, when you are defining the linear transformation L, you index j from 1 to m instead of 1 to n.

These are pretty small details, but I think they could cause confusion for someone unfamiliar with the material!

Thanks! –Jason

[Thanks for the corrections – T.]

20 March, 2010 at 10:31 am

Matt

Dear Professor Tao:

On page 520 of Analysis II, it states that the Fourier Transform is a map from functions in C(R/Z;R). Should the codomain of the functions be the complex numbers? It seems inconsistent with the rest of the surrounding definitions and theorems inasmuch as it only applies to functions with a real codomain. Thank you very much.

Matt

24 June, 2010 at 1:31 pm

Guilherme

Dear Prof. Tao,

I really liked your book. The contents of both volumes are very very good.
I was thrilled. I was thinking omg, this will be my favourite analysis book.

But I saw that you place the proof of some theorems and lemmas as exercises. I really have huge difficulties to learn from books which use this style.

When this happens, it seems that suddenly the author of the book “abandoned me”. And then I get lost. Of course, it a good habit of doing the proofs by myself, but, in the end, I am not sure my proof is good/rigorous or not.

But, the organization and the writing style are amazing !

Best,
Guilherme

24 June, 2010 at 1:41 pm

Terence Tao

This is very much an intentional choice on my part (see page xiii of the preface). There are a lot of important subtleties in real analysis (e.g. the distinction between pointwise convergence and uniform convergence, to give just one example), and the only way to really appreciate this is to chew through some of the foundational results regarding these subtleties. There are some parts of mathematics where it is sufficient to just know the main results and their proofs, as opposed to deriving them with one’s own mental resources, but in my view analysis is not one of these subjects.

24 June, 2010 at 2:09 pm

Guilherme

Dear Prof. Tao,

It is very nice to have this feedback about learning methods from you. And it is indeed true that I learn better when I put some effort in trying to do the proof on my own before seeing the proof from the actual book.

When this happens, I appreciate much more the proof and understand better the overall(big picture) strategy.

On the other hand, what happens if the student doesnt know how to make the proof, or if the student doesnt know if the his proof is wrong ? Don’t you think that the student will miss an important part of the material ?

ps: I am not making this questions with the intention to criticize. Mainly, I am making this questions in order to improve my learning methods too.
I always get frustrated when I realized that the author puts the proofs in the exercises. Maybe it is time for a change. :)

Best,
Guilherme

24 June, 2010 at 2:30 pm

Terence Tao

A student that is unable to make proofs, or to verify whether a given proof is correct, will have difficulty with a subject as theory-intensive as real analysis regardless of whether the textbook contains proofs of its theorems or not, particularly when one reaches the point where one needs a result not already contained in the textbook.

In the case of my own textbook, in most cases when a proof of a theorem is set as an exercise, a proof of a related or similar result is provided in the text as a model; the ability to adapt an argument from one context to another is another important skill in analysis that is difficult to pick up by any means other than direct practice.

10 October, 2012 at 4:54 am

Luqing Ye

I am in the contrary.I learn mathematics by proving theorems.Though in this way I learn math slowly,i quite enjoy it ,that’s enough.In learning Prof.Tao’s book(and other people’s book),I just prove everything I encounter,it is very often that I spend several days proving a difficult theorem,that doesn’t matter,time is used to waste.When a theorem is proved by me,and I believe that my proof is correct,I will write down my proof,and put it in my personal blog(If a theorem is proved by me,I will not read author’s proof,I have no patient absorbing other people’s idea).Why I write down my proof ?Because I like to record things,if it is not recorded,several weeks later,it may go like wind,on the other hand,if a theorem is proved by me ,maybe the proof is original,original things must be recorded.Record my own proof also helps me to review,maybe in the future I will forget what I learned(that’s very often for me),then I will review my proof of the theorems in that field,then I will quickly recall nearly everything,by the way,in the process of reviewing,I will read my proof seriously the second times to check whether it has some mistake in it .That’s how I learn mathematics,maybe in the future I should adjust my method a bit,it is very tired to do every myself.

31 March, 2013 at 9:29 am

Misha Shvartsman

Following that logic, it would not make sense to have any homework exercise that requires a proof. And, indeed, these days browsing internet you probably can find any proof you want, but chance that you learn anything will not be great :)

24 June, 2010 at 6:24 pm

Tom

Dear Terry,

Can one use matrices and linear algebra to explain both pointwise and uniform convergence?

9 August, 2010 at 5:17 am

Bart Kleijngeld

It’s only a minor correction, but they all count, right? On page 536, example 17.1.7 you write $x \in \mathbb{R}$ , while it should be $c \in \mathbb{R}$ .

Thanks for your great book,
Bart

[Corrected, thanks – T.]

7 September, 2010 at 7:30 pm

Student

Dear Prof. Tao, I am having trouble with 13.2.10 in analysis 2. I have to show that
dx((x,y), (x0, y0)) is greater than dx((x,y0), (x0, y0)) (holding one variable constant). It seems rather obvious, but I cannot seem to get it.

7 September, 2010 at 7:36 pm

Terence Tao

That’s not quite the right approach. I recommend that you write down in full (using epsilons and deltas) what it means for the two-variable function $(x,y) \mapsto f(x,y)$ to be continuous at $(x_0,y_0)$ , and what it means for the function $x \mapsto f(x,y_0)$ to be continuous at $x_0$ , taking particular care to get all the quantifiers (“for all”, “there exists”) correct and in the right order. Once you have both definitions down side by side, it should become clear how to proceed.

14 September, 2010 at 3:11 pm

Juan

Hello Dr. Tao,

I am having trouble with 13.4.6 in your book, Analysis II. I have been trying to do a proof by contradiction, assuming that the union of an arbitrary number of connected sets is in fact disconnected, but I am stuck.

Here is what I have:

Call the arbitrary union of the connected sets, E.
Assume that there exist open, disjoint, nonempty sets A, B, such that A union B = E.

Then A = complement of B.

Ea (the alpha’th set in the collection of connected sets). Assume Ea is not contained in A nor B (rather it is “split up” among the two sets). Thus there exists a subset of A, call it D, and a subset of B, call it E, s.t. D union E = Ea. But Ea is connected, so D and E are either nondisjoint, or one of the sets is not open. That is where I am stuck. What do I do from there? I assume that I want to show that for all alpha, Ea must be contained in either A or B, so then we get a contradiction, but I do not know.

14 September, 2010 at 7:56 pm

Terence Tao

If A is open in E, then $D := A \cap E_\alpha$ will be open in $E_\alpha$ .

It may help to isolate a common point $x_0$ of the intersection $\bigcap_\alpha E_\alpha$ , which is non-empty by hypothesis. Note that $x_0$ must belong to exactly one of A or B.

4 November, 2010 at 9:04 am

Kent Van Vels

Hello Dr. Tao,

I am going through your analysis book. I may have found a typo in the second edition. In exercise 16.5.4, should the formula be

$\widehat{f'(n)} = i n \widehat{f}(n)$ be changed to

$\widehat{f'(n)} = 2\pi i n \widehat{f}$ ?

Thanks,

Kent

[Added, thanks – T.]

4 November, 2010 at 9:06 am

Kent Van Vels

I mean, of course,

$\widehat{f'}(n) = 2\pi i n \widehat{f}(n)$ .

13 November, 2010 at 11:10 am

Kent Van Vels

I have a question the second edition of Analysis II.
For the proof of Proposition 14.1.5 (exercise 14.1.2)

Can you explain where my reasoning goes wrong in showing that the third statement is logically distinct from the first statement?

If we consider the function $f:\mathbb{R}\to \mathbb{R}$ defined as $f(x) = 0$ for $x\not =0$ , and $f(x) = 1$ for $x=0$ . Then
$\lim_{x\to 0; x\in \mathbb{R}} f(x) = 0$ .
Now, let $V=(-1/2,1/2)$ , an open set which contains $0$ . Now, every open set that contains $0$ will contain an interval of the form $(-a,b)$ where $a,b>0$ . But $f((-a,b) \cap \mathbb{R}) = \{0,1\} \not \subset (-1/2,1/2)= V$

Do we need to assume that $f$ is continuous or that $x_{0}$ is a limit point (instead of an adherent point)?

Thanks,

Kent Van Vels

13 November, 2010 at 12:14 pm

Terence Tao

For your specific function, the limit $\lim_{x \to 0; x \in {\bf R}} f(x)$ does not exist, because you are not excluding 0 from the range of possible values of x. (This notational convention may be a bit different from what is seen in some other texts; see Remark 14.1.2.)

13 November, 2010 at 1:11 pm

Kent Van Vels

Thank you. That clears up my confusion.

23 November, 2010 at 2:55 pm

Anonymous

Prof. Tao,

After studying the Chapter 17 Several variable differential calculus in Analysis vol.2 , I am curious about whether there are high order derivatives for the function $f: R^n\rightarrow R^m$ . But a further consideration seems to tell me that the high order derivative is not the transformation as $L_{Df}$ any more. It seems that one has to define the “second derivative” for the transformation $L_{Df}:R^n\rightarrow R^{m\times n}$ . But it seems strange.

Could you explain about this? If they do exist, what’s the topic there in the larger picture? Or is this concept meaningless?

23 November, 2010 at 8:12 pm

Terence Tao

These higher order derivatives are best described either as multilinear maps or as tensors. For instance, the second derivative is a bilinear map that is essentially the Hessian. The third and higher derivatives can also be defined, but they are less pleasant to work with, basically because multilinear algebra is significantly less pleasant than linear algebra in many ways. (Also, when working on a curved domain, the underlying curvature of the space begins to assert its influence much more strenuously on the third and higher derivatives than on the first and second ones, leading to further technical complications.)

30 November, 2010 at 11:09 am

Kent Van Vels

In Lemma 18.4.8 (the proof of countable additivity of Lebesgue measure), The third sentence that begins “Let $A$ be any other measurable set…”. Is it ok to only consider measurable sets $A$ to show that $E$ measurable?

[Oops, $A$ should be an arbitrary set rather than a measurable one. I’ve added an appropriate erratum – T.]

1 December, 2010 at 2:50 pm

Kent Van Vels

Can you explain where my reasoning goes wrong in showing a
counterexample to Exercise 13.5.5?

Let $X$ be the set $\{0,1\}\subset \mathbb{N}$ , with the $0\prec 1$ . Then $X$ is totally ordered. But for
$x=0$ (an element of $X$ ) we can’t find an $a\in X$
such that $a\prec x$ . This shows
that $X$ isn’t open, and so $(X,\mathcal{F})$ isn’t a
topology.

1 December, 2010 at 4:16 pm

Terence Tao

Oops, one also needs to allow for half-rays and the whole space in addition to intervals. I’ve added an appropriate erratum.

1 December, 2010 at 7:32 pm

Kent Van Vels

Should Exercise 13.5.6 and Exercise 13.5.7 be changed to read that the topologies in question are non-Hausdorff? It seems impossible for two disjoint open sets to exist.

[Corrected, thanks – T.]

2 December, 2010 at 11:00 am

KVV

In Proposition 12.4.12 it says

“Let $(X,d)$ be a metric space…” should this be changed to read

“Let $(X,d)$ be a complete metric space?

Thanks.

2 December, 2010 at 11:17 am

Terence Tao

X needs to be complete for part (b), but this is not needed for part (a).

4 December, 2010 at 11:01 pm

Kent Van Vels

Thanks. That clears up my misunderstanding.

Kent

4 December, 2010 at 12:59 pm

KVV

In exercise 12.5.12 there is a hint that reads in part,

“…useless here since that only applies to Euclidean spaces with the Euclidean metric.” Should that be changed to indicate that the Heine-Borel theorem also applies to Euclidean spaces equipped with the sup-norm and taxi-cab metric?

Thanks,

Kent

[Corrected, thanks – T.]

12 December, 2010 at 10:52 am

KVV

On page 515 in the second edition there might be two small typos.

There is math display that reads, $latex\cdots \sum_{j=\in J}\cdots$, I don’t think the equals sign is required.

Also, in the definition of the outer measure, there is a sum from $j=1$ to $\infty$ . Should this be a sum over $j\in J$ ?

Thanks,

Kent

[Corrected, thanks – T.]

13 December, 2010 at 8:24 am

KVV

Hello,

On page 545, in the statement of Corollary 19.2.11, there is a sequence of functions. Do we have to assume that each function in the sequence is measurable?

Thanks,

Kent

[Corrected, thanks – T.]

29 December, 2010 at 1:07 pm

KVV

I think there is a small typo in the statement of Exercise 15.2.8. For part (e), there is a sum whose summand is $d_m (x-b)^n$ , I think it should probably be $d_m (x-b)^m$ .

Thanks.

[Corrected, thanks – T.]

17 January, 2011 at 3:22 pm

KVV

I think there are two small typos in the statement of exercise 14.1.6 on page 396, (second edition).

First, there is a display $\lim_{y\to y_0; y\in f(E)} g(x) = z_0$ , I think it should probably be $\lim_{y\to y_0;y\in f(E)} g(y) = z_0$ .

Second, there is a display $\lim_{x\in x_0; x\in E} g \circ f(x_0) = z_0$ , I think it should probably be $\lim_{x\to x_0; x\in E} g\circ f(x) = z_0$

Thanks

[Corrected, thanks – T.]

13 March, 2011 at 10:56 pm

Biswa

Dear Prof. Tao,

On Page 478, just before Definition 15.2.4, “…and differentiable on (a-r, a+r) for some $a\in{\mathbb R}$” should be “…and differentiable on (a-r, a+r) for some $r>0$.

Thank you.
Biswaranjan Behera

[Corrected, thanks – T.]

15 March, 2011 at 3:26 am

Biswa

P 485, $s_N$ (third display) and $s_0$ (next line) should be
$S_N$ and $S_0$ . Also, should it be “by convention
$S_0=-D$ in place of “in particular $S_0=-D$ ?

[Corrected, thanks. The way summation has been defined in the text, an empty summation is automatically zero without the need for any explicit further convention. -T.]

30 April, 2011 at 1:52 pm

Anonymous

I am confused by the exercise 14.1.1. Why “the limit $lim_{x\to x_0;x\in E}f(x)$ exists iff the limit $lim_{x\to x_0;x\in E\setminus \{x_0\}}f(x)$ AND is equal to $f(x_0)$ ?” This is true only when $f:E\to Y$ is continuous, isn’t it?

30 April, 2011 at 11:55 pm

Terence Tao

No, continuity is not required for this fact. (Both the hypothesis and the conclusion of the exercise, though, are equivalent to $f$ being continuous at $x_0$ .)

1 May, 2011 at 8:43 pm

Anonymous

Dear Prof. Tao,

I am not sure if this is the right place to put the following question here.

In the very beginning of the book, Chapter 1 of volume I, you introduce what is real analysis and point out that it is the theoretical foundation “which underlies calculus”. However, I didn’t find that you talk about the “multivariable calculus” in you real analysis series courses, from math131AB to math 245ABC.

Actually, you did talk about the “several variable differential calculus” in Chapter 17 in this volume. But why there is no “multivariable integration” here? Is it because it is enough for the theory which you have talked about in your series books(Analysis I,II; An introduction to measure theory;An epsilon of room, Vol I) to build up the “multivariable integration”, or because it is beyond the scope of this course?

Since I’d like to build up the system of knowledge about real analysis myself, I follow the series of your books. But it puzzles me that I am even not able to find the “multivariable integration” in your books.

1 May, 2011 at 9:34 pm

Terence Tao

See Chapter 19.

1 May, 2011 at 9:55 pm

Anonymous

I did not put my question in the right way. I think I am talking about the [multiple integral](http://en.wikipedia.org/wiki/Multiple_integral) and the [vector analysis](http://en.wikipedia.org/wiki/Vector_calculus). In the one dimensional case, you talked about the Riemann integral in Chapter 11 and finally the Lebesgue integration in Chapter 19. In the n-dimensional case, you introduce the Lebesgue integration in ${\bf R}^n$ in Chapter 19. So there is no “Riemann integral” but the Lebesgue integration in the multi-dimensional case? What’s the theoretical counterpart of multiple integral in the vector analysis in your book?

2 May, 2011 at 7:08 am

Terence Tao

See Section 19.5.

Generalising the one-dimensional Riemann integral as set up in Chapter 11 to higher dimensions would be a good exercise for a student who has gone through that chapter. (I explicitly set that exercise, incidentally, as Exercise 1.1.26 of my measure theory book.) However, I feel that this is something that a good student can work out on his or her own, after understanding the core topics covered in texts such as my own (as well as seeing a lower-division treatment to higher-dimensional integration).

4 June, 2011 at 3:36 pm

Anonymous

I find that you didn’t deal with the “Local maxima, local minima” in the multivariate case in these two volumes. Is this intentional? Will you recommend any references which deal with this topic rigorously?

12 June, 2011 at 8:08 am

Anonymous

Errata to the second edition, ‘formula does not parse’

[Corrected, thanks – T.]

12 June, 2011 at 9:23 am

Anonymous

Dear Prof. Tao,
I am confused with the definition of partial derivative(p.546 Definition 17.3.7). Is it a special case of the “directional derivatives”? If it is, why the constrain for $t$ in the definition is $t\to 0;t\neq 0$ instead of $t\to 0;t>0$ which appears in the definition of directional derivative? If it is not, how can you conclude from Lemma 17.3.5 that $\frac{\partial f}{\partial x_j}(x_0)=f'(x_0)e_j$ ?

12 June, 2011 at 10:49 am

Terence Tao

The partial derivative $\frac{\partial f}{\partial x_j}$ is related to the two directional derivatives $D_{e_j} f$ and $D_{-e_j} f$ in that whenever $D_{e_j} f(x)$ and $-D_{-e_j} f(x)$ exist and are equal to each other, then the partial derivative $\frac{\partial f}{\partial x_j}(x)$ also exists and is equal to these two quantities, and conversely (this follows from Proposition 9.5.3).

12 June, 2011 at 7:15 pm

Anonymous

Dear Prof. Tao,

For the Exercise 17.3.3, by definition, $D_vf(0.0)=\frac{v_1^3}{v_1^2+v_2^2}$ where $v=(v_1,v_2)$ . and ${\partial}{\partial x}f(0,0)=1$ , ${\partial}{\partial y}f(0,0)=0$ . Then by your remark in p.547, $D_vf(0,0)=\sum_jv_j\frac{\partial f}{\partial x_j}(0,0)$ , $f$ is not differentiable at $(0,0)$ . Is the proof supposed to be like this? Or do you have any hint for this?

[This will work, yes – T.]

12 June, 2011 at 7:19 pm

Anonymous

Dear Prof. Tao,
I didn’t see any connection between Exercise 17.3.1 and Exercise 17.1.3. (You said that “this will be similar to Exercise 17.1.3″)Typo?

[Sorry, this should be Exercise 17.2.1. -T.]

15 June, 2011 at 9:20 am

Chris

Dear Professor Tao

I had another confusion that require clarification.

1) In your above errata, you mention, Pg 478: Just before definition, ………. should be ………..

My comment: The actual page is 427 but I can’t find the one that need to be replace. Can you please justify this?

2) As to your statement on Pg 485, ………….

My Comment: The correction can be made by me because the page number is wrong and I can’t identify where is that part in the text that need to be replaced.

3) Pg 548 you mention that m should be n. Can you clarify further as to what should be replaced in the text.

4) Pg 593 & Pg 625 .<— For this errata, there is no such pages in your second edition book. Can you justify which should be corrected in the second edition as I can't locate what need to be corrected.

[All errata for the first edition should already be corrected for the second edition – T.]

16 June, 2011 at 7:12 pm

Anonymous

In the section 17.8: The implicit function theorem, p.568, the notation is a little confusing for me. In the first paragraph, $f:{\bf R}^n\to{\bf R}$ gives rise to a graph $\{(x,f(x)):x\in{\bf R}^n\}$ in ${\bf R}^{n+1}$ . And in the second paragraph, $f$ is used in “a hypersurface of the form $\{x\in{\bf R}^n:f(x)=0\}$ , where $f:{\bf R}^n\to {\bf R}$ is some function”. Are these two $f$ the same?? I guess the $f$ in the theorem 17.8.1 is the same as that in the second paragraph, while $g$ is the one in the first paragraph? Am I right?

[Yes, that is right; I will change the first f to a g to reduce confusion. -T.]

9 August, 2011 at 2:56 am

Anonymous

p. 563, line 10: “We remark that …if $f(x_0$ is” should be “We remark that …if $f'(x_0$ is”.

Biswa

[Added, thanks – T.]

9 August, 2011 at 9:48 pm

Anonymous

p. 560, Lemma 17.6.6, Can one apply the contraction mapping theorem to F as it is defined on B(0, r)?

[Oops; I’ve adjusted the argument there accordingly. -T.]

9 August, 2011 at 9:59 pm

Anonymous

p. 567, line -2, (0, 1) should be (-1, 0).

[Added, thanks – T.]

12 August, 2011 at 9:21 pm

Anonymous

p.563, line 17, ‘ $f^{-1}(y)={\tilde f}^{-1}(y+f(x_0))$ ‘ should be ‘ $f^{-1}(y)={\tilde f}^{-1}(y-f(x_0))$ ‘..

p.570, second line of the last paragraph, ‘U is open and contains $(y_1, \dots,y_{n-1},0)$ ‘ should be ‘U is open and contains $(y_1, \dots,y_{n-1})$ ‘.

Biswa

8 February, 2012 at 9:02 pm

Anonymous

Can anyone show me if this integral $\int_{\pi}^{3\pi/2}\frac{1}{x^2 \sin^{3/2}(x)}dx$ is finite or not?

Thanks

5 May, 2012 at 5:50 am

ugroh

Terence, two comments on Analysis 2:

p.25 Theorem 15.1.6. (d): Should not the formula for $f'$ starts with
$n=1$ instead of $n=0$ ? This would be more consistent with Prop. 15.2.6. and the formula there.

p.427, 3rd line from the top: … differentiable on .. for some $r \in R$ instead of $a \in R$ ?

Ulrich

[Corrections added, although I will have to recheck the page numbers on the errata list the next time I have access to a physical copy of the book, as some of the numbers seem to be incorrect.]

6 May, 2012 at 9:58 pm

ugroh

Sorry, it should be p. 425 not p. 25
Ulrich

25 July, 2012 at 5:24 am

Gökhan Güçlü

Dear Professor,

In the proof of Theorem 12.5.8 Case 1, in the second paragraph (As before, we know that there exists…) all the sequences should be subsequences

[Correction added, thanks -T.]

10 October, 2012 at 6:35 am

Luqing Ye

Dear Prof.Tao,

At the beginning of section 17.7,you say:

“We recall the inverse function theorem in single variable calculus(Theorem 10.4.2), which asserts that if a function $f:\mathbf{R} \to \mathbf{R}$ is invertible, differentiable, and $f'(x_0)$ is non-zero, then $f^{-1}(x)$ isdifferentiable at $f(x_0)$ , and $(f^{-1})'(f(x_0))=\frac{1}{f'(x_0)}$ .”

I think maybe you lost a condition……I think the condition “ $f^{-1}$ is continuous at $f(x_0)$ ” should be added.Theorem 10.4.2 has that condition.

10 October, 2012 at 10:37 am

Luqing Ye

Oops,It is my fault.At the beginning of section 17.7,it is $f:\mathbb{R}\to \mathbb{R}$ ,in this case, $f^{-1}:\mathbb{R}\to \mathbb{R}$ have to be continuous automatically.But in theorem 10.4.2,it is $f:X\to Y$ , $X,Y$ are arbitrary point sets. …..Oh my god,awesome mathematics……Sorry for my fault.

20 October, 2012 at 4:59 am

Matthis Lehmkühler

Dear Prof. Tao,

in Example 18.2.9. (p. 520 Second Edition, Hardcover), there is a little mistake in the middle part of the inequation: You have to replace $\sum_{q\in\mathbf{Q}} m*(\mathbf{Q})$ by $\sum_{q\in\mathbf{Q}} m*(\{q\})$ as mentioned in the above text.

25 October, 2012 at 11:16 am

Matthis Lehmkühler

Dear Prof. Tao,

in the proof of Theorem 17.5.4 (p. 495f. Second Edition, Hardcover) it sais that one has to proof that $a=a'$ whereby $a$ is defined to be $a:=\frac{\partial}{\partial x_j}\frac{\partial f}{\partial x_i}(0)$ and $a'$ should denote the quantity $a':=\frac{\partial}{\partial x_j}\frac{\partial f}{\partial x_i}(0)$ . Of course one has to replace $a':=\frac{\partial}{\partial x_j}\frac{\partial f}{\partial x_i}(0)$ by $a':=\frac{\partial}{\partial x_i}\frac{\partial f}{\partial x_j}(0)$ .

[Added, thanks – T.]

1 November, 2012 at 9:11 am

Anonymous

Dear Prof. Tao,
In exercise 14.1.2(prove Proposition 14.1.5), I think statement (d) isn’t logical
equivalent with (a) when $x_0\in E$ and $f(x_0)\neq L$ . such as:
$x_0=0$ , $E=[-1,1]$ , $f(x)=0(x\neq 0)$ , $f(0)=1$ , $L=0$ , then $g(x)=0$ is continuous at $0$ , but the limit $\lim_{x\rightarrow 0;x\in[-1,1]}f(x)$ does no exist.

[Yes, you’re right; I’ve added an erratum accordingly. -T.]

9 January, 2013 at 5:37 am

Jack

The power series can be generalized to complex functions. Can the Weierstrass approximation also true in complex analysis, say every complex continuous function in a closed disc in ${\bf C}$ can be uniformly approximated by polynomials? (I don’t see an obvious way to translate the proof in terms of complex functions.)

9 January, 2013 at 9:31 am

Terence Tao

If one considers polynomials of both $z$ and $\overline{z}$ , then one can recover the Weierstrass approximation theorem in the complex case; but if one restricts attention to polynomials of $z$ only, then the Weierstrass approximation theorem fails (one can approximate functions that are analytic in a neighbourhood of one’s domain, but not general continuous functions). See Section 3 of https://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/ for more discussion.

20 January, 2013 at 4:35 pm

Jack

Can one get from Theorem 15.1.6 (c) that $\sum_{n=0}^{\infty}c_n(x-a)^n$ converge uniformly to $f$ on the open interval $(a-R,a+R)$ ?

I saw several theorems about uniform convergence on a compact set or a closed bounded interval. And in complex analysis, there are also similar results such as “if $f$ is a holomorphic function in a disc $D$ , then it has a power series expansion that converges uniformly on every compact set in $D$ “.

Why is the uniform convergence on compact sets so special?

21 January, 2013 at 8:22 am

Jack

After some thought, I managed to come up with $\sum_{n=0}x^n$ as a counterexample. And the uniform convergence on compact sets is discussed in detail in 245b Notes 11(https://terrytao.wordpress.com/2009/02/21/245b-notes-11-the-strong-and-weak-topologies/). [Since I’m note able to edit the original questions (as people usually do in MO), I have to put the further words in a new comment box. Merging them together might be good though.]

The 245ABC series notes seem to be very complete regarding fundamental real analysis that one should be able to find an answer in them regarding basic questions of real analysis, although the author didn’t mention what’s “not” in detail (or not covered) in these notes :-)

7 March, 2013 at 5:55 am

Luqing Ye

Dear Prof.Tao,

In exercise 13.3.5,you say:Let $(X,d_X)$ be a metric space,and let $f:X\to \mathbf{R}$ and $g:X\to\mathbf{R}$ be uniformly continous functions,show that the direct sum $f\bigoplus g:X\to\mathbf{R}^2$ defined by $f\bigoplus g(x):=(f(x),g(x))$ is uniformly continous.

I think it is necessary to equip $\mathbf{R}^2$ with the Euclidean metric and $\mathbf{R}$ with the standard metric.Or maybe you have already said it earlier but I didn’t notice…

7 March, 2013 at 6:34 am

Luqing Ye

Oh my god,in Example 12.1.4 you have said without special remark we equip $\mathbf{R}$ with standard metric.But I can’t find where you said without special remark we equip $\mathbf{R}^2$ with Euclidean metric…

[A remark to this effect has been added to the errata, thanks. -T.]

9 March, 2013 at 10:41 pm

Luqing Ye

Dear Professor Tao,

Exercise 13.5.1 asks me to prove that when $\#X>1$ ,then the trival topology can not be obtained by placing a metric on $X$ .But I think this conclusion is not correct.

For example,if $X=\{p,q\}$ ,and $p\neq q$ ,then if we place a discret metric $d$ on X,then $d(p,p)=d(q,q)=0,d(p,q)=d(q,p)=1$ ,let $\tau=\{\emptyset,X\}$ ,then obviously $(X,\tau)$ is a trival topolggy.

So I obtained a trival topology by placing a metric on X ！

It is almost true that my argument might be wrong,the key point is what does “obtained ” mean..

10 March, 2013 at 12:51 pm

Terence Tao

“obtained” is in the sense in the paragraph after Definition 13.5.1, namely one needs to take the topology to be the collection of sets that are open with respect to the metric. In your example, the open sets with respect to the discrete metric are $\{ \emptyset, \{p\}, \{q\}, X \}$ , which is not the trivial topology.

10 March, 2013 at 9:04 pm

Luqing Ye

Thanks a lot.By the way,in your errata you say “In Exercises 13.5.6 and 13.5.7, “Hausdorff” should be “not Hausdorff”.”

This makes me amazing!Because yesterday I proved they are Hausdorff!

So it make me think what is a proof……I can prove what I believe,even though that is wrong…And I can’t prove what I don’t believe…

Oh my god…Prove rigourously is easier said than done.

10 March, 2013 at 3:48 am

Luqing Ye

Dear Prof.Tao,

An errata of your errata.

p. 390: In Exercise 13.5.5, “there exist a, b \leq X such that the “interval” \{ y \in X: a < y < b \}” should be replaced with “there exists a set I which is an interval \{y \in X: a < y < b\} for some a, b \in X, a ray \{y \in X: a < y \} for some a \in A, the ray \{ y \in X: y < b\} for some a \in A, or the whole space X, which”.

It should be

p. 390: In Exercise 13.5.5, “there exist $a, b \leq X$ such that the “interval” $\{ y \in X: a < y < b \}$ ” should be replaced with “there exists a set I which is an interval $\{y \in X: a < y < b\}$ for some $a, b \in X$ , a ray $\{y \in X: a < y \}$ for some $a \in A$ , the ray $\{ y \in X: y < b\}$ for some $b \in A$ , or the whole space X, which”.

I hope the errata of the errata of your errata won't exist.

[Errata corrected, thanks – T.]

10 March, 2013 at 11:54 pm

Luqing Ye

Dear Professor Tao,and everyone,

Sorry I have so many comments(They are not posted casually),but I think I found a counterexample to Exercise 13.5.8.

Exercise 13.5.8 is stated as follows:

Let X be an uncountable set,and let $\infty$ be an element of X.Let $\tau$ be the collection of all subsets E in X which are either empty or are co-countable and countain $\infty$ .Show that $(X,\tau)$ is a compact topological space…

But I think $(X,\tau)$ is not neccessarily a compact topological space.

The counterexample is as follows,this counterexample make use of the ordinals.

Lets see a set of ordinal numbers which are uncountable,they are $1,2,3,4,\cdots,n,\cdots,\omega,\omega+1,\omega+2,\cdots$ (According to the well ordering principle,an uncountable sets $X$ can be well ordered and thus order isomorphic to an ordinal).

Let $\omega=\infty$ ,Now we see the subsets $P_0=\{\omega,\omega+1,\cdots,\omega+n,\cdots\}$ , $P_1=P\bigcup\{1\}$ ,$\cdots$, $P_n=P_{n-1}\bigcup \{n\}$ .

It can be easily verified that $P_0,P_1,\cdots,P_n,\cdots$ is an open covering of X,and $\forall i\in\mathbf{N}$ , $P_i$ contains the element $\infty$ ( $\omega$ ),And it is also easy to verify that a finite subsets of $\{P_0,P_1,\cdots,P_n,\cdots\}$ can not cover X.Thus X is not a compact topological space.

Maybe there are mistakes in my arguement,I will appreciate any comments.If there is no mistake in my argument,then I really find a counterexample!r

11 March, 2013 at 8:02 am

Terence Tao

Sorry, the topology given in the exercise (the neighbourhood topology of the cocountable topology at infinity) was not the one intended; it is the order topology on $\omega_1+1$ which gives the required counterexample (namely, a topology that is compact but not sequentially compact). The replacement exercise is now given in the errata.

14 March, 2013 at 3:05 am

Luqing Ye

Dear professor Tao,

I can’t prove this new exercise,because I think I find a counterexample.The difficulty is that I cannot express my counterexample very clearly by using English,but I will try that.

Firstly,I will define a term “gap”.To understand what does “gap” mean,Lets see these ordinals:

$0,1,2,\cdots,n,\cdots,\omega,\omega+1,\cdots$

We call there is a gap between a natural number and the ordinal $\omega$ .If you start from a natural number $n$ and begin to walk step by step in the positive direction,one second a step,then you will never go out of that gap.

Now I begin to introduce my counterexample .

$\omega_1+1$ is an ordinal,for an element $x$ in $\omega_1+1$ ,if $x$ is a successor ordinal,now we prove that the set $\{x\}$ is an open set in the context of the order topology ,because if $x=\infty(\omega_1+1)$ ,then $x\in \{p\in \omega_1+1:p>\omega_1\}$ ,and $\{p\in\omega_1+1:p>\omega_1\}\subseteq \{\infty\}$ ;If $x\neq\infty$ ,then $x\in\{p\in\omega_1+1:x-1<p<x+1\}$ ,and $\{p\in\omega_1+1:x-1<p<x+1\}\subseteq \{x\}$ .

So when $x$ is a successor ordinal,I just cover $x$ by the open set $\{x\}$ .

If $x$ is a limit ordinal,and there are countable infinite gaps before $x$ (Obviously,such $x$ exists),then we cover $x$ by the set $\{p\in\omega_1+1:a<p\leq x\}$ ,where $a$ is a successor ordinal,and there are countable infinite number of gaps before $a$ (Obviously,such $a$ exists).Now we prove that $\{p\in\omega_1+1:a<p\leq x\}$ is an open set that covers $x$ .For $y\in \{p\in\omega_1+1:a<p\leq x\}$ ,if $y=a$ ,then $y\in \{p\in\omega_1+1:a-1<y<a+1\}$ ,and $\{p\in\omega_1+1:a-1<y<a+1\}\subseteq \{a\}$ ;if $y=x$ , $y\in \{p\in\omega_1+1:a<y<x+1\}$ ,and $\{p\in\omega_1+1:a<y<x+1\}\subseteq \{p\in \omega_1+1:a<p\leq x\}$ .So $\{p\in\omega_1+1:a<p\leq x\}$ is an open set that covers $x$ .So when $x$ is a limit ordinal and there are countable infinite gaps before $x$ ,I just cover it by the open set $\{p\in\omega_1+1:a<p\leq x\}$ .

It is easy to verify that there are countable infinite number of limit ordinals which has the above feature(countable infinite gaps before it),so it is easy to verify that we have constructed an infinite open cover which does not have finite open cover.

Done.

Maybe my counterexample is wrong,or maybe I do not express it well(But I examine my comments several times before it is posted).I will appreciate any comments.

14 March, 2013 at 5:55 am

Luqing Ye

A supplement to my comment:

$\omega_1+1$ here is a special “ordinal”,because I regard the maximal element $\infty$ in $\omega_1+1$ as $\omega_1+1$ itself.This doesn’t matter.

14 March, 2013 at 6:59 am

Luqing Ye

No,No,I cannot do something which is contradict the axiom of regularity.My supplement is wrong.

Dear prof.Tao,I am in mess,I am a newbie to such thing as ordinals…Could you please give me some hints to your new exercise,because now I even doubt the correctness of your new exercise……I think it is not compact,I can construct a open cover which has no finite subcover.

15 March, 2013 at 12:22 am

Luqing Ye

My counterexample is wrong because of the lack of knowlege in ordinal type.After see a brief introduction to ordinals in wikipedia,now I understand where I am wrong…

20 September, 2014 at 2:29 am

Anonymous

Hi, Luqing Ye, can you please show me how to prove the new exercise? in fact i have no idea to that exercise.

20 March, 2013 at 11:42 pm

Luqing Ye

Dear Prof.Tao,

In definition 14.8.4,you say “A function R–> R is said to be compactly supported iff it is supported on some interval [a,b]”

What does “some” mean?Some means “Infinite” or “finite(more than nothing)”,Or “nothing”?

21 March, 2013 at 1:10 am

Luqing Ye

OH,maybe here “some ” means “a”,because you write “interval” instead of “intervals”.But if it means “a”,then the concept of compactly supported function is same to the concept of the supported function…

21 March, 2013 at 12:57 pm

Anonymous

I m basing my answer only on your question here, I do not have the book at hand.

But it should be understood like this:

A function R -> R is said to be compactly supported, if there exists an interval [a,b], such that supp f \subset [a,b].

21 March, 2013 at 8:01 pm

Luqing Ye

Thanks.So if a function is compactly supported,then this function is supportted,and if this function is compactly supported,then this function is also supported.

Then why we need the definition of compactly supported?It is equivalent to the definition of supported……

[The definition on the book is following:
Let [a,b] be an interval.A function f:R–>R is said to be supported on [a,b] iff f(x)=0 for all $x\not\in [a,b]$ .We say that f is compactly supported iff it is supported on some interval [a,b].

]

22 March, 2013 at 1:04 am

Anonymous

Hello, the first definition is not that of a function being supported, but of a function being supported on a concrete interval [a,b]!

The definition says: Let [a,b] be an interval.A function f:R–>R is said to be !! supported on [a,b] !! …

I hope that clearifies your question!

22 March, 2013 at 1:36 am

Shen Zeng

Just in case my answer was too short, here is an example:

Take for example the characteristic function of [0,1], i.e. f(x) = 1 iff x in [0,1] and 0 otherwise.

This function is supported on [0,1]. It is supported also on [0,2], but not on [2,3]. Lastly, it is compactly supported, because it is supported for at least one interval [a,b], i.e. a=0 and b=1 or a=0 and b=2. This is because it f is supported on [0,1] and f is supported on [0,2].

I hope the difference and connections between these definitions became clear. If not, feel free to ask again!

22 March, 2013 at 1:55 am

Luqing Ye

Thanks for your explanation and your example,now I understand all…

【supported on some interval [a,b]】means 【there exists a interval [a,b] such that f is supported on [a,b]】

Maybe I have to go back to primary school……

27 March, 2013 at 11:49 pm

Luqing Ye

Hi,Prof.Tao,

In exercise 15.2.8,(d),you say: $|d_m|\leq C(s-\varepsilon)^m$ .But in (b),you say $|c_n|\leq C(r-\varepsilon)^{-n}$ .

I can’t prove (d),so I wonder whether $C(s-\varepsilon)^m$ should be changed into $C(s-\varepsilon)^{-m}$ …in that way,(b) and (d) would have the same form and will looks more comfortable…..

[Errata added, thanks – T.]

28 March, 2013 at 9:45 pm

Luqing Ye

Here is my proof.I can’t proceed any further.

———————————————————–
$\displaystyle |d_m|=|\sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!} (b-a)^{n-m} c_n|\leq\sum_{n=m}^{\infty}|\frac{n!}{m!(n-m)!}(b-a)^{n-m}||c_n|\leq\sum_{n=m}^{\infty}C|\frac{n!}{m!(n-m)!}(b-a)^{n-m}(r-\varepsilon)^{-n}|=C\sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}|b-a|^{n-m}(r-\varepsilon)^{-n}$

According to Exercise 15.2.7,

$\displaystyle \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}|b-a|^{n-m}(r-\varepsilon)^{-n}=\frac{r-\varepsilon}{(r-\varepsilon-|b-a|)^{m+1}}$

$\displaystyle |d_m|\leq C \frac{r-\varepsilon}{(r-\varepsilon-|b-a|)^{m+1}}=C \frac{r-\varepsilon}{r-\varepsilon-|b-a|}(r-\varepsilon-|b-a|)^{-m}$

We know that $r-|b-a|\geq s$ ,so $r-|b-a|-\delta=s$ , $\delta\geq 0$ .So

$\displaystyle |d_m|\leq C \frac{r-\varepsilon}{r-\varepsilon-|b-a|}(s+\delta-\varepsilon)^{-m}$ .

——————————————————–

So I doubt that you should let $s$ be as large as possible,i.e,let $s=r-|a-b|$ ,so that $\delta=0$ ,then I proved it.If $s$ is not big enough,I can’t prove this result…..

In fact,I think it doesn’t matter that we let $s=r-|b-a|$ .

2 April, 2013 at 7:37 am

Terence Tao

$(s+\delta-\varepsilon)^{-m} \leq (s-\varepsilon)^{-m}$ .

2 April, 2013 at 7:38 am

Luqing Ye

Hi Mr.Tao,

In Exercise 15.7.2,I think “Show that there exists a c>0 such that f(y) is non-zero whenever 0<|x-y|<c” should be “Show that there exists a c>0 such that f(y) is non-zero whenever 0<|x_0-y|<c”.

Sorry that my comment is not always right,so it needs your wisdom to judge which is right and which is wrong….

[Errata added, thanks – T.]

8 April, 2013 at 10:02 pm

Luqing Ye

Dear Prof.Tao,

This is not an errata,but a personal advice.In lemma 16.4.6,you make use of Fejer kernel to construct an $(\varepsilon,\delta)$ approximation to the identity.Why Fejer kernel?I think a long time to find the geometric meaning of the Fejer kernel,but the geometric meaning is rather complex,I make use of a sophiscated but wrong model of planet motion found by ancient astronomers to give Fejer kernel a geometric explanation(A circle A_2 moves around a circle A_1,and a circle A_3 moves around the circle A_2,etc…).

So I think the Fejer kernel approach is rather unintuitive to a begginer in Fourier analysis.Here is an intuitive approach:

Consider the function $f_n(x)=\cos^n \frac{\pi}{2}x,x\in [0,1]$ .When $n$ become larger and larger, $f_n(x)$ will become thiner and thiner,and at last will become a $(\varepsilon,\delta)$ approximation to the identity(In order to do this ,we have to multiply a constant $c$ to f_n(x),so that $\int_{[0,1]}cf_n(x)dx=1$ ).And it is easy to verify by using algebra of trigonometrics like $2\cos^2x-1=\cos 2x$ ,we can change $f_n(x)$ into trigonometric polynomials,that is done!

9 April, 2013 at 12:17 am

Luqing Ye

Oh!This method fails!In this method,I only construct an $(\varepsilon,\delta)$ approximation to the identity,not an $(\varepsilon,\delta)$ periodic approximation to the identity!(But by this wrong method,I can prove that a series of trigonometric polynomials can approximate uniformly to a continuous function on an interval.)

But even though this method is wrong,I think I will be right only after some minor corrections.Instead of $f_n(x)=\cos^n \frac{\pi}{2}x$ ,I need to construct a trigonometric function,this function is always non-negative,and at the place of $x=1$ ,this function should become large again.

Then let this function replace $\cos \frac{\pi}{2}x$ ,then I think that will be OK.

12 April, 2013 at 3:24 am

ugroh

Terence, on Page 452 (Analysis II), Exercise 15.7.6 you are writing
“.. be a non-zero complex real number ..”. I guess the “real” can be ignored.
Regards
Ulrich

17 April, 2013 at 8:27 am

Luqing Ye

Dear Prof.Tao,

I think the definition of derivative in this book (and in Spivak’s “calculus on manifold”,etc.) is inconsistent.

When we say “A differentiable function $f:\mathbb{R}^m\to \mathbf{R}^n$ has derivative $L$ at point $x_0\in \mathbf{R}^m$ “,we regard $L$ as a linear map from $\mathbf{R}^m$ to $\mathbf{R}^n$.

But in the case of $m=n=1$ ,we regard $L$ as a real number.A real number is a real number,not a linear map,isn’t it?(Though a real number multiplying a stuff means a linear map )

It is very similar to such conditions,that is,the distinction between the matrix and the correspoinding linear map.Maybe somebody regard matrix as a linear map(I don’t know whether “somebody” exists or not,I just guess),but I regard matrix as a matrix, $m$ columns, $n$ rows.Only when a matrix multiplying a stuff forms a linear map…

17 April, 2013 at 9:07 am

Terence Tao

In one dimension, there is a canonical isomorphism that identifies each real number $L$ with the associated dilation map $x \mapsto Lx$ on the reals, so one customarily “abuses notation” by identifying the two (somewhat analogously to how one identifies natural numbers with a subset of the integers, or the rationals as a subset of the reals, etc.).

In higher dimensions, the corresponding identification between matrices and linear transformations is dependent on the choice of basis, which is why it is important to keep the two concepts separate. But in one dimension the identification is basis-independent, and there is little harm in conflating the two concepts (or, for that matter, with identifying $1 \times 1$ matrices with scalars).

17 April, 2013 at 9:34 am

Luqing Ye

Dear Prof Tao,

I just now say “Hmm,this comment is wrong again”.After seeing your reply,It seems that my comment is not wrong again,because in your book is

$\displaystyle \frac{f(x)-f(x_0)-L(x-x_0)}{|x-x_0|}$ ,In spivak’s book is $\displaystyle \frac{f(x)-f(x_0)-h(x)}{|x-x_0|}$ .

Now I understand,thanks for your reply!

17 April, 2013 at 9:11 am

Luqing Ye

Hmm…This comment is wrong again…In fact,when $m=n=1$ ,the linear transformation is $h(x):f'(x_0)(x-x_0)$ ,as illustrated in Spivak’s book…

23 May, 2013 at 10:16 pm

BWW

Dear Professor Tao,

Should the second inequality in (12.1) on page 392 read $d_{l_1}(x,y)\leq nd_{l^2}(x,y)$ instead of $d_{l_1}(x,y)\leq \sqrt{n}d_{l^2}(x,y)$ ? I think $d_{l^1}((.4,.4),(0,0))=.8\geq \sqrt{2}d_{l^2}((.4,.4),(0,0))$ is a counterexample to the original inequality.

24 May, 2013 at 2:08 am

Luqing Ye

I don’t think so.Cauchy’s inequality can’t be wrong.

In your counterexample, $d_{l^1}((.4,.4),(0,0))=.8= \sqrt{2}d_{l^2}((.4,.4),(0,0))$ ,which does not conflict $d_{l_1}(x,y)\leq \sqrt{n}d_{l^2}(x,y)$ .

13 July, 2013 at 4:15 am

Andre

Exercise 14.1.1. Is the requirement that the if you are assume that the limits are equal to $f(x_0)$, then the equivalence holds ? So you require that $f$ is continuous ?

13 July, 2013 at 7:12 am

Terence Tao

Not quite. The first part of the question asks to show, under the hypotheses stated, that “ $\lim_{x \to x_0; x \in E} f(x)$ exists” is true if and only if “ $\lim_{x \to x_0; x \in E \backslash \{x_0\}} f(x)$ exists AND this limit is equal to $f(x_0)$ ” is true. The second part of the question asks to show that “ $\lim_{x \to x_0; x \in E} f(x)$ exists” implies that “ $\lim_{x \to x_0; x \in E} f(x)$ is equal to $f(x_0)$ “. In neither case is continuity of $f$ assumed, although if any of the above premises are true then it is indeed the case that $f$ is necessarily continuous at $x_0$ when restricted to $E$ .

13 July, 2013 at 9:32 am

Andre

Thank’s a lot.

5 October, 2013 at 11:41 am

Anonymous

Do we really need $f$ being countinuously differentiable to recover the pointwise convergence of the Fourier series to $f$ as you stated in Remark 16.5.2? A local result states that if $f$ being an integrable function on the circle is differentiable at a point $\theta_0$ , then the Fourier series converges to $f$ at that point. Is it because we need global pointwise convergence that the countinuity of the derivative is needed?

5 October, 2013 at 12:50 pm

Terence Tao

Hmm, you’re right; continuous differentiability may be relaxed to just differentiability for the purposes of this remark.

5 October, 2013 at 4:49 pm

Jack

Can continuous differentiability of $f$ actually give absolute and uniform convergence of the Fourier series?

5 October, 2013 at 5:44 pm

Terence Tao

For continuously differentiable functions, I believe that one has conditional uniform convergence, but not absolute convergence even at a pointwise level, let alone uniformly. (A random Fourier series $\sum_{n=1}^\infty \pm \frac{1}{n} \cos(2\pi n x)$ will almost surely give a counterexample for the latter.)

5 October, 2013 at 7:20 pm

Jack

I don’t understand what you mean by “conditional uniform convergence”. There is a remark in Stein and Shakarchi’s Fourier Analysis which says that the Fourier series of $f$ converges absolutely, assuming only that $f$ converges absolutely (and hence uniformly to $f$ ) if $f$ satisfies a holder condition of order $\alpha>1/2$ . Where do I get things wrong?

5 October, 2013 at 7:36 pm

Terence Tao

Ah, you’re right; continuous differentiability (or Holder continuity with exponent greater than 1/2) does indeed give absolute uniform convergence and not just conditional uniform convergence. (I had misplaced a factor of $\sqrt{n}$ in my previous calculation.)

14 December, 2013 at 5:40 pm

Dear Prof Tao,
I have some problem for exercise 18.4.3,how I get the result frome 18.4.2?
Thank you!

23 February, 2014 at 3:45 am

Martin Scorsese

Dear Prof. Tao,

is there some reason you define elements of $\mathbb{R}^n$ to be row vectors instead of column vectors ? In every linear algebra book that I looked (e.g. Langs “Linear Algebra” oder Axlers “Linear Algebra”) and also in analysis books, like Rudins “Principles of Mathematical Analysis”, these are (in the case of the analysis books usually only implicitly) treated as column vectors (since no transposition is involved when multipliying them with matrices). Therefore I think there have to be other good reasons, which are unkown to me, to treat them as row vector if one is thereby accepting the additional hassle of having to transpose them before multiplying them against matrices…

Also regarding linear algebra you wrote on the 17 April, 2013 at 9:07 am, that there is a canonical isomorphism that maps every number $L$ to the map $x\mapsto Lx$. Why is this map “canonical” ? For one-dimensional vector spaces $V$ the correspondence $V\leftrightarrow V^*$ isn’t canonical since it depends on a chosen basis of $V$. Therefore also the correspondence $\mathbb{R}\leftrightarrow \mathbb{R}^*$ which you mentioned doesn’t seem to be canonical.

23 February, 2014 at 8:39 am

Terence Tao

Unfortunately, the most natural conventions in linear algebra have evolved to be the opposite of the conventions for the rest of mathematics. For instance, elements of a Cartesian product $X \times Y$ are invariably written as horizontal ordered pairs $(x,y)$ rather than vertical ones $\begin{pmatrix} x \\ y \end{pmatrix}$ ; similarly, functions $f$ of two variables $x,y$ are denoted $f(x,y)$ rather than $f \begin{pmatrix} x \\ y \end{pmatrix}$ , and so forth. This is probably for historical typesetting reasons rather than for any mathematical reason (and, as one can see here, column vectors still waste a lot of whitespace in modern typesetting), but still, if one insists on ${\bf R}^n$ as column vectors, then basic statements such as ${\bf R}^{n+m} = {\bf R}^n \times {\bf R}^m$ require some modification. So the compromise has been to use column vectors in situations in which linear algebra is the dominant focus, but to use row vectors if it is not (and to convert between the two by transposes if the focus varies within the text).

Ultimately, the source of the conflict is the fact that we do not use the “reverse Polish” notation of putting operations after their inputs, instead of before their inputs, e.g. $(x,y)f$ instead of $f(x,y)$ or $f \begin{pmatrix} x \\ y \end{pmatrix}$ , but it is far too much trouble to change over the entire corpus of mathematical writing (and habit) for such minor notational optimisations, as the “repaired” notation just looks too “wrong” to almost all practising mathematicians, including myself. (Though there are certainly exceptions: for instance, many mathematicians who work with group actions have a preference for right actions over left actions.)

As for your second question, ${\bf R}$ does have a canonical basis, namely the multiplicative unit 1, and hence a canonical identification with its dual. Here we are implicitly viewing ${\bf R}$ not just as a one-dimensional vector space, but as the much richer structure endowed with the structure of a field (and with many other structures besides: order structure, topological structure, measure structure, etc.).

27 February, 2014 at 2:13 am

Scorsese

I’m really sorry I have to bother you again, but I thought hard about your explanation and I can’t seem to understand these two arguments of yours:
$\bullet$ why statements like $\mathbb{R}^{m+n}=\mathbb{R}^m \times \mathbb{R}^n$ require some modifications and
$\bullet$ why reverse Polish notation would solve this conflict.

These are my concerns:
Regarding the first problem, set-theoretically $\mathbb{R}^{m+n}=\mathbb{R}^m \times \mathbb{R}^n$ , these actually means that between the set of tuples of the form $(x_1,\ldots,x_{m+n})$ and the set of tuples of the form $((x_1,\ldots,x_{m}),(x_{m+1},\ldots,x_{m+n})$ exists an (obvious) bijection – but, as you wrote at the top of pp. 73 – we neglect these minor distinctions and therefor write $\mathbb{R}^{m+n}=\mathbb{R}^m \times \mathbb{R}^n$ as if the sets were indeed the same.
But writing columnvectors instead of row vectors doesn’t change anything since then – completely analoguous to the row-vector case – we would again have a bijection

$\begin{pmatrix} x_1 \\ \vdots \\ x_{m+1} \\ \vdots \\ x_{m+n} \end{pmatrix} \mapsto \begin{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_m \end{pmatrix} \\ \begin{pmatrix} x_{m+1} \\ \vdots \\ x_{m+n} \end{pmatrix} \end{pmatrix}$
and could choose to neglect this minor difference.

Regarding the second problem: Polish notation is just a notation whereas row and column vector are objects that have a mathematical existance (since e.g. for column vectors of length 2 multiplication from the left with a $2\times 2$ matrix is defined but for row vector it isn’t). So how can a typographical change induce a mathematical change ?

27 February, 2014 at 6:11 am

Terence Tao

For the first point: one can certainly consistently use column notation for both vectors and ordered pairs as you do here, but the problem is typographical: due to the way English is written (from left to right, and then top to bottom), using vertical notation like this is very wasteful in space, does not look aesthetically appealing, and is also more cumbersome to write in formatting languages such as LaTeX. Furthermore, outside of linear algebra, ordered pairs (and n-tuples) are invariably written in row notation instead of column notation, so switching everything over to columns is a much larger change to mathematical notation than switching everything over to rows. (In an alternate universe in which the dominant language was written from top to bottom, rather than from left to right, as is the case for instance with traditional Chinese, it would make sense to use column notation throughout, but this is not the case in the universe we actually live in.)

For the second point: when one switches to reverse Polish notation, the natural association between linear operators and matrices changes. In orthodox notation, a linear transformation $T: {\bf R}^n \to {\bf R}^m$ between standard Euclidean spaces is associated to an $m \times n$ matrix $A$ by the formula $Tx = Ax$ for all $x \in {\bf R}^n$ , where elements of ${\bf R}^n$ are viewed as column vectors. In reverse Polish notation, it is instead natural to associate a linear transformation $T: {\bf R}^n \to {\bf R}^m$ to an $n \times m$ matrix $A$ by the formula $xT = xA$ for all $x \in {\bf R}^n$ , where elements of ${\bf R}^n$ are now viewed as row vectors. Note also that the dimensions of the matrix are now ordered in the same way as we order the domain and range of the function, i.e. with the domain dimension $n$ to the left of the range dimension $m$ . This is related to how the operation of applying a function $f$ and then a function $g$ becomes $fg$ or $f \circ g$ (or $f g \circ$ ) in reverse Polish notation, as opposed to $gf$ or $g \circ f$ in orthodox notation: reverse Polish notation makes function composition consistent with the visual operation of concatenation of arrows. (In orthodox notation, one can “repair” this issue by reversing the arrows and placing the domain to the right of the range, e.g. $f: Y \leftarrow X$ (or $Y \leftarrow X: f$ ) instead of $f: X \to Y$ , but this usually deviates too far from orthodox practice to be widely used.)

28 February, 2014 at 1:28 pm

Scorsese

You are great! Thank you so much!

10 March, 2014 at 2:15 pm

Jack

As I run through the book, I don’t see the topic “Differentiation under the integral sign“.

In the preface, you point out that careless interchanging integrals (limits, limits and derivatives, derivatives, limits and integrals) might lead to big mistakes. You’ve also talked about “There’s more to mathematics than rigour and proofs“.

I’ve seen two of your blog posts mentioned in the beginning that we can ignore the issue of “differentiation under the integral sign” for the sake of “formal level of analysis”:

“Noether’s theorem, and the conservation laws for the Euler equations“:

Throughout this post, we will work only at the formal level of analysis, ignoring issues of convergence of integrals, justifying differentiation under the integral sign

“Some notes on Bakry-Emery theory“:

we will ignore all technical issues of regularity and decay, and always assume that the solutions to equations such as (1) have all the regularity and decay in order to justify all formal operations such as the chain rule, integration by parts, or differentiation under the integral sign.

You once said that in the “post-rigorous” stage (of one’s mathematical education) when the emphasis is on applications, intuition, and the “big picture”, one should be able to convert all (formal) calculations into a rigorous argument whenever required.

Is there a “simple” rule of thumb that one can justify “differentiation under the integral sign”? (It seems that this is very “easy” according to those two blog posts.) To what extend can one safely do this sort of formal discussion?

As I see from lots of (should I say “all”?) your advanced discussions in analysis, almost none of the exact statement in the book (except some really big theorem like Weierstrass’s Theorem) would be mentioned. How should one understand this phenomenon?

10 March, 2014 at 3:24 pm

Terence Tao

Writing a derivative as a limit of Newton quotients, the problem of interchanging derivatives and integrals is more or less equivalent to a problem of interchanging limits and integrals. So, for instance, if the Newton quotients are all uniformly dominated by an absolutely integrable function, the dominated convergence theorem will allow one to justify the interchange. (This is for instance the case if the functions involved are smooth and compactly supported in all parameters; one can weaken these hypotheses further of course.)

As for your last question, the short answer is that most of modern analysis is not so much about theorems, but rather about techniques and ideas; see for instance Tim Gowers’ “The two cultures of mathematics” for further discussion of this point.

24 March, 2014 at 12:00 pm

Anonymous

Terence,

I am currently an undegraduate who is going to do Real Analysis next semester. While Rudin is the most used book, several reports claim that it is not “user-friendly”, whatever that means in a mathematical context. In case I feel the same way; does your books on analysis serve as an alternative? In particular, are your books suitable for Real Analysis? (I am not sure what to take from “Honors Analysis”, as honors-classes are not practiced in Norway.)

24 June, 2014 at 5:56 am

Martin Scorsese

Dear Prof. Tao,

I’m don’t really understand the point of most of the exercises (for example exercise 15.6.11 or 15.6.13) on complex numbers from pp. 502 (in the first edition).
These exercises mostly concern purely topological notions of $\mathbb{C}$ , which, by the set-theoretic definition of $\mathbb{C}$ as $\mathbb{R}^2$ together with a suitable multiplication, have already been established and proved earlier in the text.
Therefore the proof of these exercises is just a “proof by notation” and their proof has an somewhat disturbing, “tautological” feeling to it – also because they are phrased in a way as if we were proving a new result.
For example in 15.6.11 $f$ is just the identity, written in a fancy way as $f(a,b)=(a,b)=(a,0)+(0,1)\cdot (0,b)=a+ib$ using complex multiplication, and thus trivially a homeomorphism and in 15.6.13 it seems to me one doesn’t really need 15.6.11, since set-theoretically $\mathbb{C}=\mathbb{R}^2$ and thus the proof of the Heine-Borel theorem from earlier on can be used verbatim.
I’m sure you had your reasons, but wouldn’t a general remark, that the topological aspects of $\mathbb{C}$ are already all established because they have been proved earlier for $\mathbb{R}^2$ , have been sufficient ?

24 June, 2014 at 8:55 am

Terence Tao

Hmm, this is a good point. I guess the issue was blurred somewhat by overloading the $(a,b)$ notation to represent both the ordered pairs used in the Cartesian product, and the formal version of $a+bi$ in the definition of the complex numbers; it’s slightly better conceptually to think of ${\bf R}^2$ as a model (or as a coordinate chart) of ${\bf C}$ , rather than being set-theoretically identical to ${\bf C}$ . The distinction between ${\bf C}$ and ${\bf R}^2$ can be important if one wishes to consider other ways to coordinatise the complex plane, or if one wishes to exploit general linear symmetries on the plane ${\bf R}^2$ that do not preserve the complex structure. Unfortunately I did not set up the notation properly to emphasise this distinction (to do so, I would have used something like $a \oplus b \textbf{i}$ rather than $(a,b)$ to represent a formal complex number, although that looks somewhat ugly and perhaps would have caused additional confusion).

(There is also one tiny piece of mathematical content to Exercise 15.6.11, which is to check that the Euclidean metric on ${\bf R}^2$ agrees with the complex metric on ${\bf C}$ , although this is trivially verifiable in one line, of course.)

These sorts of questions become more relevant if one is studying more complicated spaces, e.g. the Riemann sphere ${\bf CP}^1$ , which has two coordinate charts by ${\bf C}$ (or by ${\bf R}^2$ ), in which case establishing basic topological properties of this space (e.g. completeness or connectedness) requires slightly more effort, but I think I agree with you that in this context a remark that ${\bf C}$ is identical as a metric space to ${\bf R}^2$ is simpler. (This does reinforce the slightly incorrect notion that ${\bf C}$ “is” ${\bf R}^2$ as a mathematical space, not just as a metric space, but in the grand scheme of things this is a fairly harmless notion to hold.)

26 August, 2014 at 12:03 am

Anonymous

Dear Prof. Tao, i have a confusion:
in Proposition 12.2.15 :
(d) Any singleton set {xo}, where xo E X, is automatically
closed.

but in 12.3 Relative topology, you show us a example:
however if instead
one uses the discrete metric ddisc, then { 1} is now an open set

26 August, 2014 at 10:23 pm

Terence Tao

It is possible for a set to be simultaneously open and closed (such sets are known as clopen sets).

26 August, 2014 at 11:05 pm

Anonymous

the intuitive is ingrained. i must forget those common sense. thank you for your patience.

26 August, 2014 at 10:55 pm

Anonymous

It is possible for a set to be simultaneously
open and closed, if it has no boundary. so no confusion.

3 September, 2014 at 11:16 pm

Anonymous

Exercise 12.5.8. Let ( $X, d_{l^1}$ ) be the metric space from Exercise 12.1.15.
For each natural number n, let $e^{(n)} = {{e_j}^{(n)}}_{j = 0}^{\infty}$ be the sequence in $X$
such that ${e_j}^{(n)} := 1$ when n = j and ${e_j}^{(n)} := 0$ when $n \neq j$ Show that
the set $\{e^{(n)} : n \in N\}$ is a closed and bounded subset of $X$ , but is not
compact. (This is despite the fact that (( $X, d_{l^1}$ ) is even a complete metric
space ¨C a fact which we will not prove here. The problem is that not
that $X$ is incomplete, but rather that it is ¡°infinite-dimensional¡± , in a
sense that we will not discuss here.)

in fact , i have spend several days to prove that : ( $X, d_{l^1}$ ) is a complete metric space. but i can¡¯t got it. can you please show me some clues? thank you.

25 September, 2014 at 2:36 pm

Anonymous

Look in any good book about Functionalanalysis like Yoshida or Schechter …

3 September, 2014 at 11:29 pm

Anonymous

You can adopt the method e.g. in Schechter, Principles of Functional Analysis, 1.3, Example 3 to prove this.

4 September, 2014 at 3:34 am

Anonymous

I got it, thank you.

10 September, 2014 at 11:11 pm

Anonymous

Dear Prof. Tao
in Exercise 13.3.6. Show that the addition function (x,y) $\mapsto$ x+y and the
subtraction function (x, y) $\mapsto$ x – y are uniformly continuous from $R^2$
to R, but the multiplication function (x, y) $\mapsto$ xy is not. Conclude that
if f : X $\to$ R and g : X $\to$ R are uniformly continuous functions on
a metric space (X, d), then f + g : X $\to$ Rand f – g : X $\to$ Rare
also uniformly continuous. Give an example to show that fg: X $\to$ R
need not be uniformly continuous. What is the situation for max(f, g),
min(f,g), f/g, and cf for a real number c?

I want to proof max(f,g) is uniformly continuous.
in fact , for any $\epsilon/2$ , there exist $\delta$ , when d(x,x’) < $\delta$ , $ d(f(x),f(x')) < \epsilon/2$ and $ d(g(x),g(x')) < \epsilon/2$.

in situation max(f,g), when d(x,x') = g(x) and f(x’) >= g(x’) (or g(x) >= f(x) and g(x’) >= f(x’)) then max(f,g) = f, it’s obvious.

however, it f(x) > g(x) and g(x’) > f(x’) (or g(x) > f(x) and f(x’) > g(x’)) , i want to found one point $x_0$ , such that $f(x_0) = g(x_0)$ , $d(x,x_0) < \delta$ and $d(x',x_0) < \delta$ ,then based on Triangle inequality ,we got it.

the key point is How to found that $x_0$ ?, there seems NOT have something similar to Intermediate value theorem which need connected.

OR max(f,g) is NOT uniformly continuous?

11 September, 2014 at 1:05 am

Anonymous

i think this should work:

In fact , for any $\varepsilon$ , there exist $\delta$ , when $d_x(x,x') < \delta$ , we have: $d(f(x),f(x')) = |f(x) - f(x')| < \varepsilon$ and $d(g(x),g(x')) = |g(x) - g(x')| < \varepsilon$ .

In situation max(f,g):
Case1: $f(x) \ge g(x)$ and $f(x') \ge g(x')$ ( $g(x) \ge f(x)$ and $g(x') \ge f(x')$ is similar ) then $d(max(x),max(x')) = d(f(x),f(x')) g(x)$ and $g(x') > f(x')$ ( $g(x) > f(x)$ and $f(x') > g(x')$ is similar). based on inequality formulas on R, we have : $f(x) - f(x') > f(x) - g(x') > g(x) - g(x')$ , with $d(f(x),f(x')) = |f(x) - f(x')| < \varepsilon$ and $d(g(x),g(x')) = |g(x) - g(x')| < \varepsilon$ . we have $d(max(x),max(x')) = d(f(x),g(x')) = |f(x) - g(x')| < \varepsilon$ .

so max(f,g) is uniformly continuous.

18 September, 2014 at 7:26 pm

Anonymous

Dear Prof. Tao, in Exercise 13.5.5. Given any totally ordered set $X$ with order relation $\leq$ ,
declare a set $V \subset X$ to be open if for every $x \in V$ there exist $a, b \in X$
such that the “interval” $\{y \in X : a < y < b\}$ contains $x$ and is contained
in $V$ . Let $\mathcal{F}$ be the set of all open subsets of X. Show that $(X,\mathcal{F})$ is
a topology.

I have a problem with this exercise:

We know $X \not \subset X$ ,so $X \notin \mathcal{F}$ , and so $(X,\mathcal{F})$ is NOT a topology? ,so $V \subset X$ shoulde be $V \subseteq X$ ?

even with $V \subseteq X$ hold, later for $R ^{*}$ case , then $+\infty \in R ^{*}$ ,BUT, there DOES NOT exist $a, b \in R ^{*}$
such that the "interval" $\{y \in R ^{*} : a < y < b\}$ contains $+\infty$ , $R ^{*} \notin \mathcal{F}$ , so $(R ^{*},\mathcal{F})$ is NOT a topology?

19 September, 2014 at 8:10 am

Terence Tao

Thanks for the correction. (As for the second issue, see the errata for this question for the second edition, which has been corrected in the third edition.)

23 September, 2014 at 8:34 am

ua46f

Dear Prof. Tao, in your new Exercise 13.5.8 : Show that there exists an uncountable well-ordered set $\omega_1+1$ that has a maximal element $\infty$ , and such that the initial segments $\{ x \in \omega_1+1: x < y \}$ are countable for all $y \in \omega_1+1 \backslash \{\infty\}$ . If we give $\omega_1+1$ the order topology (Exercise 13.5.5), show that $\omega_1+1$ is compact; however, show that not every sequence has a convergent subsequence.

After follow your hint , we can get well order set : $\omega_1+1$ ,apply Transfinite induction to $\omega_1+1$ , we can get: $\{ x \in \omega_1+1: x < y \}$ are countable for all $y \in \omega_1+1 \backslash \{\infty\}$ , so a initial segment it is.

my problem is whether the condition "countable" is necessary for $\omega_1+1$ to be compact?

i have this proof which don't use the "countable" condition:

let $\omega_1+1 = \bigcup _ {c \in I} V_c$ , $I$ is some infinite set. then $\infty \in \omega_1+1$ ,there must exist some $V_{c^m}$ which contain $\infty$ , we can found $V_{c^{m-1}}$ which have : $min(V_{c^{m-1}}) < min(V_{c^{m}})$ and for every $s, min(V_{c^{m-1}}) < s < min(V_{c^{m}}) , s \in V_{c^{m-1}}$ , and so forth, then $min(V_{c^{m}}), min(V_{c^{m-1}}), \dots$ is a strick descendant sequence ,so which must be finite. so $\omega_1+1 = \bigcup _ {0 \leq n \leq m ; c_n \in I} V_{c^n}$ , a finite cover.

24 September, 2014 at 11:32 am

Terence Tao

Yes, one can replace $\omega_1$ by any other successor ordinal and still obtain topological compactness in the order topology. (Whether sequential compactness holds in that case is another question, though.)

25 September, 2014 at 6:24 am

ua46f

Dear Prof. Tao, in Exercise 13.5.13. Generalize Corollary 12.5.9 to compact sets in a topological space.

Corollary 12.5.9. Let $(X,\;d)$ be a metric space, and let $K_1,\;K_2,\;K_3$
be a sequence of non-empty compact subsets of $X$ such that
$K_1 \; \supset \;K_2 \; \supset K_3\; \supset \; \dots$
Then the intersection $\bigcap _{n=1} ^{\infty} K_n$ is non-empty.

I think we needs to assume as an additional hypothesis that the topological space is Hausdorff.

If we don’t have this hypothesis, i just don’t know how to prove $K_n$ is closed.

[Correction added, thanks -T.]

25 September, 2014 at 2:32 pm

Anonymous

Metrische Spaceshuttle are Hausdorff spaces.

25 September, 2014 at 2:34 pm

Anonymous

Sorry, metric spaces are ….

28 September, 2014 at 5:26 pm

José Antonio Lara Benítez

Dear Prof. Tao I think there is a typo in the proposition 18.3.3 instead of $m^{*}(\sum _{q\in J}q+E )= \sum _{q\in J}m^{*}(q+E)$ should be $m^{*}(\bigcup_{q\in J}q+E )= \sum _{q\in J}m^{*}(q+E)$ . In the same way at the end of the proof should be $\bigcup_{q\in J}q+E$ is a subset of $X$

[Correction added, thanks -T.]

1 November, 2014 at 2:06 pm

José Antonio Lara Benítez

Dear Professor Tao. In the Corollary 18.4.7 p. 592 I think is necessary to assume that $m(A)<\infty$ in order to avoid some indeterminate form.

[Correction added, thanks – T.]

3 November, 2014 at 5:13 pm

José Antonio Lara Benítez

Dear professor Tao. I think there is a problem with the hint in the exercise 19.1.3 p. 608. Because as the $f_n$ ‘s are defined doesn’t necessarily implies that are simple as is required in the lemma 19.1.5 p. 603.

3 November, 2014 at 11:03 pm

Terence Tao

Actually, I believe the $f_n$ defined in the hint are indeed simple (they can only take finitely many values, namely the multiples of $2^{-n}$ between $0$ and $2^n$ ).

4 November, 2014 at 12:25 am

Anonymous

in Lemma 16.4.6 , $\sum_{n=0}^{N-1}e_n(x) = \frac{e_N - e_0}{e_1 - e_0} = \frac{e^{\pi i(N-1)x} sin(\pi Nx)}{sin(\pi x)}$ then $F_N(x) = \frac{{sin(\pi Nx)}^2}{N{sin(\pi x)}^2}$

SHOULD BE $\sum_{n=0}^{N-1}e_n(x) = \frac{e_N - e_0}{e_1 - e_0} = \frac{e^{\frac{\pi i (N-1)x}{2}} sin(\frac{\pi N x}{2})}{sin(\frac{\pi x}{2})}$ then $F_N(x) = \frac{{sin(\frac{\pi Nx}{2})}^2}{N{sin(\frac{\pi x}{2})}^2}$ ?

4 November, 2014 at 12:09 pm

Terence Tao

No; note that $e_n(x)$ is equal to $e^{2\pi i nx}$ , not $e^{\pi i nx}$ .

4 November, 2014 at 5:33 pm

Anonymous

yes, i miss the definition . thank you for your time.

4 November, 2014 at 1:16 am

Anonymous

one more question: if this is case ,the condition $0 < \delta < 1/2$ , $\delta \leq |x| \leq 1 - \delta$ SHOULD BE $0 < \delta < 1$ , $\delta \leq |x| \leq 1$ this is same to Definition 14.8.6

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