A key theme in real analysis is that of studying general functions or by first approximating them by “simpler” or “nicer” functions. But the precise class of “simple” or “nice” functions may vary from context to context. In measure theory, for instance, it is common to approximate measurable functions by indicator functions or simple functions. But in other parts of analysis, it is often more convenient to approximate rough functions by continuous or smooth functions (perhaps with compact support, or some other decay condition), or by functions in some algebraic class, such as the class of polynomials or trigonometric polynomials.

In order to approximate rough functions by more continuous ones, one of course needs tools that can generate continuous functions with some specified behaviour. The two basic tools for this are Urysohn’s lemma, which approximates indicator functions by continuous functions, and the Tietze extension theorem, which extends continuous functions on a subdomain to continuous functions on a larger domain. An important consequence of these theorems is the Riesz representation theorem for linear functionals on the space of compactly supported continuous functions, which describes such functionals in terms of Radon measures.

Sometimes, approximation by continuous functions is not enough; one must approximate continuous functions in turn by an even smoother class of functions. A useful tool in this regard is the Stone-Weierstrass theorem, that generalises the classical Weierstrass approximation theorem to more general algebras of functions.

As an application of this theory (and of many of the results accumulated in previous lecture notes), we will present (in an optional section) the commutative Gelfand-Neimark theorem classifying all commutative unital -algebras.

** — 1. Urysohn’s lemma — **

Let be a topological space. An indicator function in this space will not typically be a continuous function (indeed, if is connected, this only happens when is the empty set or the whole set). Nevertheless, for certain topological spaces, it is possible to approximate an indicator function by a continuous function, as follows.

Lemma 1 (Urysohn’s lemma)Let be a topological space. Then the following are equivalent:

- (i) Every pair of disjoint closed sets in can be separated by disjoint open neighbourhoods , .
- (ii) For every closed set in and every open neighbourhood of , there exists an open set and a closed set such that .
- (iii) For every pair of disjoint closed sets in , there exists a continuous function which equals on and on .
- (iv) For every closed set in and every open neighbourhood of , there exists a continuous function such that for all .

A topological space which obeys any (and hence all) of (i-iv) is known as a normal space; definition (i) is traditionally taken to be the standard definition of normality. We will give some examples of normal spaces shortly.

*Proof:* The equivalence of (iii) and (iv) is clear, as the complement of a closed set is an open set and vice versa. The equivalence of (i) and (ii) follows similarly.

To deduce (i) from (iii), let be disjoint closed sets, let be as in (iii), and let be the open sets and .

The only remaining task is to deduce (iv) from (ii). Suppose we have a closed set and an open set with . Applying (ii), we can find an open set and a closed set such that

Applying (ii) two more times, we can find more open sets and closed sets such that

Iterating this process, we can construct open sets and closed sets for every dyadic rational in such that for all , and for any .

If we now define , where ranges over dyadic rationals between and , and with the convention that the empty set has sup and inf , one easily verifies that the sets and are open for every real number , and so is continuous as required.

The definition of normality is very similar to the Hausdorff property, which separates pairs of points instead of closed sets. Indeed, if every point in is closed (a property known as the property), then normality clearly implies the Hausdorff property. The converse is not always true, but (as the term suggests) in practice most topological spaces one works with in real analysis are normal. For instance:

Exercise 1Show that every metric space is normal.

Exercise 2Let be a Hausdorff space.

- Show that a compact subset of and a point disjoint from that set can always be separated by open neighbourhoods.
- Show that a pair of disjoint compact subsets of can always be separated by open neighbourhoods.
- Show that every compact Hausdorff space is normal.

Exercise 3Let be the real line with the usual topology , and let be the topology on generated by and the set consisting only of the rationals ; in other words, is the coarsest refinement of the usual topology that makes the set of rationals an open set. Show that is Hausdorff, with every point closed, but is not normal.

The above example was a simple but somewhat artificial example of a non-normal space. One can create more “natural” examples of non-normal Hausdorff spaces (with every point closed), but establishing non-normality becomes more difficult. The following example is due to Stone.

Exercise 4Let be the space of natural number-valued tuples , endowed with the product topology (i.e. the topology of pointwise convergence).

- Show that is Hausdorff, and every point is closed.
- For , let be the set of all tuples such that for all outside of a countable set, and such that is injective on this finite set (i.e. there do not exist distinct such that ). Show that are disjoint and closed.
- Show that given any open neighbourhood of , there exists disjoint finite subsets of and an injective function such that for any , any such that for all and is identically on , lies in .
- Show that any open neighbourhood of and any open neighbourhood of necessarily intersect, and so is not normal.
- Conclude that with the product topology is not normal.

The property of being normal is a topological one, thus if one topological space is normal, then any other topological space homeomorphic to it is also normal. However, (unlike, say, the Hausdorff property), the property of being normal is not preserved under passage to subspaces:

Exercise 5Given an example of a subspace of a normal space which is not normal. (Hint: use Exercise 4, possibly after replacing with a homeomorphic equivalent.)

Let be the space of real continuous compactly supported functions on . Urysohn’s lemma generates a large number of useful elements of , in the case when is locally compact Hausdorff:

Exercise 6Let be a locally compact Hausdorff space, let be a compact set, and let be an open neighbourhood of . Show that there exists such that for all . (Hint:First use the local compactness of to find a neighbourhood of with compact closure; then restrict to this neighbourhood. The closure of is now a compact set; restrict everything to this set, at which point the space becomes normal.)

One consequence of this exercise is that tends to be dense in many other function spaces. We give an important example here:

Definition 2 (Radon measure)Let be a locally compact Hausdorff space that is also -compact, and let be the Borel -algebra. An (unsigned) Radon measure is a unsigned measure with the following properties:

- (Local finiteness) For any compact subset of , is finite.
- (Outer regularity) For any Borel set of , .
- (Inner regularity) For any Borel set of , .

Example 1Lebesgue measure on is a Radon measure, as is any absolutely continuous unsigned measure , where . More generally, if is Radon and is a finite unsigned measure which is absolutely continuous with respect to , then is Radon. On the other hand, counting measure on is not Radon (it is not locally finite). It is possible to define Radon measures on Hausdorff spaces that are not -compact or locally compact, but the theory is more subtle and will not be considered here. We will study Radon measures more thoroughly in the next section.

Proposition 3Let be a locally compact Hausdorff space that is also -compact, and let be a Radon measure on . Then for any , is a dense subset in (real-valued) . In other words, every element of can be expressed as a limit (in ) of continuous functions of compact support.

*Proof:* Since continuous functions of compact support are bounded, and compact sets have finite measure, we see that is a subspace of . We need to show that the closure of this space contains all of .

Let be a Borel set of finite measure. Applying inner and outer regularity, we can find a sequence of compact sets and open sets such that . Applying Exercise 6, we can then find such that . In particular, this implies (by the squeeze theorem) that converges in to (here we use the finiteness of ); thus lies in for any measurable set . By linearity, all simple functions lie in ; taking closures, we see that any function lies in , as desired.

Of course, the real-valued version of the above proposition immediately implies a complex-valued analogue. On the other hand, the claim fails when :

Exercise 7Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Show that the closure of in is , the space of continuous real-valued functions which vanish at infinity (i.e. for every there exists a compact set such that for all ). Thus, in general, is not dense in .

Thus we see that the norm is strong enough to preserve continuity in the limit, whereas the norms are (locally) weaker and permit discontinuous functions to be approximated by continuous ones.

Another important consequence of Urysohn’s lemma is the Tietze extension theorem:

Theorem 4 (Tietze extension theorem)Let be a normal topological space, let be a bounded interval, let be a closed subset of , and let be a continuous function. Then there exists a continuous function which extends , i.e. for all .

*Proof:* It suffices to find an continuous extension taking values in the real line rather than in , since one can then replace by (note that min and max are continuous operations).

Let be the restriction map . This is clearly a continuous linear map; our task is to show that it is surjective, i.e. to find a solution to the equation for each . We do this by the standard analysis trick of getting an approximate solution to first, and then using iteration to boost the approximate solution to an exact solution.

Let have sup norm , thus takes values in . To solve the problem , we approximate by . By Urysohn’s lemma, we can find a continuous function such that on the closed set and on the closed set . Now, is not quite equal to ; but observe from construction that has sup norm .

Scaling this fact, we conclude that, given any , we can find a decomposition , where and .

Starting with any , we can now iterate this construction to express for all , where and . As is a Banach space, we see that converges absolutely to some limit , and that , as desired.

Remark 1Observe that Urysohn’s lemma can be viewed the special case of the Tietze extension theorem when is the union of two disjoint closed sets, and is equal to on one of these sets and equal to on the other.

Remark 2One can extend the Tietze extension theorem to finite-dimensional vector spaces: if is a closed subset of a normal vector space and is bounded and continuous, then one has a bounded continuous extension . Indeed, one simply applies the Tietze extension theorem to each component of separately. However, if the range space is replaced by a space with a non-trivial topology, then there can be topological obstructions to continuous extension. For instance, a map from a two-point set into a topological space is always continuous, but can be extended to a continuous map if and only if and lie in the same path-connected component of . Similarly, if is a map from the unit circle into a topological space , then a continuous extension from to exists if and only if the closed curve is contractible to a point in . These sorts of questions require the machinery of algebraic topology to answer them properly, and are beyond the scope of this course.

There are analogues for the Tietze extension theorem in some other categories of functions. For instance, in the Lipschitz category, we have

Exercise 8Let be a metric space, let be a subset of , and let be a Lipschitz continuous map with some Lipschitz constant (thus for all ). Show that there exists an extension of which is Lipschitz continuous with the same Lipschitz constant . (Hint:A “greedy” algorithm will work here: pick to be as large as one can get away with (or as small as one can get away with.))

One can also remove the requirement that the function be bounded in the Tietze extension theorem:

Exercise 9Let be a normal topological space, let be a closed subset of , and let be a continuous map (not necessarily bounded). Then there exists an extension of which is still continuous. (Hint: first “compress” to be bounded by working with, say, (other choices are possible), and apply the usual Tietze extension theorem. There will be some sets in which one cannot invert the compression function, but one can deal with this by a further appeal to Urysohn’s lemma to damp the extension out on such sets.)

There is also a locally compact Hausdorff version of the Tietze extension theorem:

Exercise 10Let be locally compact Hausdorff, let be compact, and let . Then there exists which extends .

Proposition 3 shows that measurable functions in can be approximated by continuous functions of compact support (cf. Littlewood’s second principle). Another approximation result in a similar spirit is Lusin’s theorem:

Theorem 5 (Lusin’s theorem)Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Let be a measurable function supported on a set of finite measure, and let . Then there exists which agrees with outside of a set of measure at most .

*Proof:* Observe that as is finite everywhere, it is bounded outside of a set of arbitrarily small measure. Thus we may assume without loss of generality that is bounded. Similarly, as is -compact (or by inner regularity), the support of differs from a compact set by a set of arbitrarily small measure; so we may assume that is also supported on a compact set . By Exercise 10, it then suffices to show that is continuous on the complement of an open set of arbitrarily small measure; by outer regularity, we may delete the adjective “open” from the preceding sentence.

As is bounded and compactly supported, lies in for every , and using Proposition 3 and Chebyshev’s inequality, it is not hard to find, for each , a function which differs from by at most outside of a set of measure at most (say). In particular, converges uniformly to outside of a set of measure at most , and is therefore continuous outside this set. The claim follows.

Another very useful application of Urysohn’s lemma is to create partitions of unity.

Lemma 6 (Partitions of unity)Let be a normal topological space, and let be a collection of closed sets that cover . For each , let be an open neighbourhood of , which are finitely overlapping in the sense that each has a neighbourhood that intersects at most finitely many of the . Then there exists a continuous function supported on for each such that for all .If is locally compact Hausdorff instead of normal, and the are compact, then one can take the to be compactly supported.

*Proof:* Suppose first that is normal. By Urysohn’s lemma, one can find a continuous function for each which is supported on and equals on the closed set . Observe that the function is well-defined, continuous and bounded below by . The claim then follows by setting .

The final claim follows by using Exercise 6 instead of Urysohn’s lemma.

Exercise 11Let be a topological space. A function is said to beupper semi-continuousif is open for all real , andlower semi-continuousif is open for all real .

- Show that an indicator function is upper semi-continuous if and only if is closed, and lower semi-continuous if and only if is open.
- If is normal and Hausdorff, show that a function is upper semi-continuous if and only if for all , and lower semi-continuous if and only if for all , where we write if for all .

** — 2. The Riesz representation theorem — **

Let be a locally compact Hausdorff space which is also -compact. In Definition 2 we defined the notion of a Radon measure. Such measures are quite common in real analysis. For instance, we have the following result.

Theorem 7Let be a non-negative finite Borel measure on a compact metric space . Then is a Radon measure.

*Proof:* As is finite, it is locally finite, so it suffices to show inner and outer regularity. Let be the collection of all Borel subsets of such that

It will then suffice to show that every Borel set lies in (note that as is compact, a subset of is closed if and only if it is compact).

Clearly contains the empty set and the whole set , and is closed under complements. It is also closed under finite unions and intersections. Indeed, given two sets , we can find a sequences , of closed sets and open sets such that and . Since

we have (by monotonicity of ) that

and similarly

and so .

One can also show that is closed under countable disjoint unions and is thus a -algebra. Indeed, given disjoint sets and , pick a closed and open such that ; then

and

for any , and the claim follows from the squeeze test.

To finish the claim it suffices to show that every open set lies in . For this it will suffice to show that is a countable union of closed sets. But as is a compact metric space, it is separable (Lemma 4 from Notes 10), and so has a countable dense subset . One then easily verifies that every point in the open set is contained in a closed ball of rational radius centred at one of the that is in turn contained in ; thus is the countable union of closed sets as desired.

This result can be extended to more general spaces than compact metric spaces, for instance to Polish spaces (provided that the measure remains finite). For instance:

Exercise 12Let be a locally compact metric space which is -compact, and let be an unsigned Borel measure which is finite on every compact set. Show that is a Radon measure.

When the assumptions of are weakened, then it is possible to find locally finite Borel measures that are not Radon measures, but they are somewhat pathological in nature.

Exercise 13Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. Define a set to be a countable union of closed sets, and a set to be a countable intersection of open sets. Show that every Borel set can be expressed as the union of an set and a null set, and as a set with a null subset removed.

If is a Radon measure on , then we can define the integral for every , since assigns every compact set a finite measure. Furthermore, is a linear functional on which is *positive* in the sense that whenever is non-negative. If we place the uniform norm on , then is continuous if and only if is finite; but we will not use continuity for now, relying instead on positivity.

The fundamentally important Riesz representation theorem for such spaces asserts that this is the *only* way to generate such linear functionals:

Theorem 8 (Riesz representation theorem for , unsigned version)Let be a locally compact Hausdorff space which is also -compact. Let be a positive linear functional. Then there exists a unique Radon measure on such that .

Remark 3The -compactness hypothesis can be dropped (after relaxing the inner regularity condition to only apply to open sets, rather than to all sets); but I will restrict attention here to the -compact case (which already covers a large fraction of the applications of this theorem) as the argument simplifies slightly.

*Proof:* We first prove the uniqueness, which is quite easy due to all the properties that Radon measures enjoy. Suppose we had two Radon measures such that ; in particular, we have

for all . Now let be a compact set, and let be an open neighbourhood of . By Exercise 6, we can find with ; applying this to (1), we conclude that

Taking suprema in and using inner regularity, we conclude that ; exchanging and we conclude that and agree on open sets; by outer regularity we then conclude that and agree on all Borel sets.

Now we prove existence, which is significantly trickier. We will initially make the simplifying assumption that is compact (so in particular ), and remove this assumption at the end of the proof.

Observe that is monotone on , thus whenever .

We would like to define the measure on Borel sets by defining . This does not work directly, because is not continuous. To get around this problem we shall begin by extending the functional to the class of bounded lower semi-continuous non-negative functions. We define for such functions by the formula

(cf. Exercise 11). This definition agrees with the existing definition of in the case when is continuous. Since is finite and is monotone, one sees that is finite (and non-negative) for all . One also easily sees that is monotone on : whenever and , and homogeneous in the sense that for all and . It is also easy to verify the super-additivity property for ; this simply reflects the linearity of on , together with the fact that if and , then .

We now complement the super-additivity property with a countably sub-additive one: if is a sequence, and is such that for all , then .

Pick a small . It will suffice to show that (say) whenever is such that , and denotes a quantity bounded in magnitude by , where is a quantity that is independent of .

Fix . For every , we can find a neighbourhood of such that for all ; we can also find such that . By shrinking if necessary, we see from the lower semicontinuity of the and that we can also ensure that for all and .

By normality, we can find open neighbourhoods of whose closure lies in . The form an open cover of . Since we are assuming to be compact, we can thus find a finite subcover of . Applying Lemma 6, we can thus find a partition of unity , where each is supported on .

Let be such that . Then we can write . If is in this sum, then , and thus (for small enough) , and hence . We can then write

and thus

(here we use the fact that and that the continuous compactly supported function is bounded). Observe that only finitely many summands are non-zero. We conclude that

(here we use that and so is finite). On the other hand, for any and any , the expression

is bounded from above by

since and , this is bounded above in turn by

We conclude that

and the sub-additivity claim follows.

Combining sub-additivity and super-additivity we see that is additive: for .

Now that we are able to integrate lower semi-continuous functions, we can start defining the Radon measure . When is open, we define by

which is well-defined and non-negative since is bounded, non-negative and lower semi-continuous. When is closed we define by complementation:

this is compatible with the definition of on open sets by additivity of , and is also non-negative. The monotonicity of implies monotonicity of : in particular, if a closed set lies in an open set , then .

Given any set , define the *outer measure*

and the *inner measure*

thus . We call a set *measurable* if . By arguing as in the proof of Theorem 7, we see that the class of measurable sets is a Boolean algebra. Next, we claim that every open set is measurable. Indeed, unwrapping all the definitions we see that

Each in this supremum is supported in some closed subset of , and from this one easily verifies that . Similarly, every closed set is measurable. We can now extend to measurable sets by declaring when is measurable; this is compatible with the previous definitions of .

Next, let be a countable sequence of disjoint measurable sets. Then for any , we can find open neighbourhoods of and closed sets in such that and . Using the sub-additivity of on , we have . Similarly, from the additivity of we have . Letting , we conclude that is measurable with . Thus the Boolean algebra of measurable sets is in fact a -algebra, and is a countably additive measure on it. From construction we also see that it is finite, outer regular, and inner regular, and therefore is a Radon measure. The only remaining thing to check is that for all . If is a finite non-negative linear combination of indicator functions of open sets, the claim is clear from the construction of and the additivity of on ; taking uniform limits, we obtain the claim for non-negative continuous functions, and then by linearity we obtain it for all functions.

This concludes the proof in the case when is compact. Now suppose that is -compact. Then we can find a partition of unity into continuous compactly supported functions , with each being contained in the support of finitely many . (Indeed, from -compactness and the locally compact Hausdorff property one can find a nested sequence of compact sets, with each in the interior of , such that . Using Exercise 6, one can find functions that equal on and are supported on ; now take and .) Observe that for all . From the compact case we see that there exists a finite Radon measure such that for all ; setting one can verify (using the monotone convergence theorem) that obeys the required properties.

Remark 4One can also construct the Radon measure using the Carátheodory extension theorem; this proof of the Riesz representation theorem can be found in many real analysis texts. A third method is to first create the space by taking the completion of with respect to the norm , and then define . It seems to me that all three proofs are about equally lengthy, and ultimately rely on the same ingredients; they all seem to have their strengths and weaknesses, and involve at least one tricky computation somewhere (in the above argument, the most tricky thing is the countable subadditivity of on lower semicontinuous functions). I have yet to find a proof of this theorem which is both clean and conceptual, and would be happy to learn of other proofs of this theorem.

Remark 5One can use the Riesz representation theorem to provide an alternate construction of Lebesgue measure, say on . Indeed, the Riemann integral already provides a positive linear functional on , which by the Riesz representation theorem must come from a Radon measure, which can be easily verified to assign the value to every interval and thus must agree with Lebesgue measure. The same approach lets one define volume measures on manifolds with a volume form.

Exercise 14Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. For any non-negative Borel measurable function , show thatand

Similarly, for any non-negative lower semi-continuous function , show that

Now we consider signed functionals on , which we now turn into a normed vector space using the uniform norm. The key lemma here is the following variant of the Jordan decomposition theorem.

Lemma 9 (Jordan decomposition for functions)Let be a (real) continuous linear functional. Then there exist positive linear functions such that .

*Proof:* For , we define

Clearly for ; one also easily verifies the homogeneity property and super-additivity property for and . On the other hand, if are such that , then we can decompose for some with and ; for instance we can take and . From this we can complement super-additivity with sub-additivity and conclude that .

Every function in can be expressed as the difference of two functions in . From the additivity and homogeneity of on we may thus extend uniquely to be a linear functional on . Since is bounded on , we see that is also. If we then define , one quickly verifies all the required properties.

Exercise 15Show that among all possible choices for the functionals appearing in the above lemma, there is a unique choice which isminimalin the sense that for any other functionals obeying the conclusions of the lemma, one has and for all .

Define a *signed Radon measure* on a -compact, locally compact Hausdorff space to be a signed Borel measure whose positive and negative variations are both Radon. It is easy to see that a signed Radon measure generates a linear functional on as before, and is continuous if is finite. We have a converse:

Exercise 16 (Riesz representation theorem, signed version)Let be a locally compact Hausdorff space which is also -compact, and let be a continuous linear functional. Then there exists a unique signed finite Radon measure such that . (Hint: combine Theorem 8 with Lemma 9.)

The space of signed finite Radon measures on is denoted , or for short.

Exercise 17Show that the space , with the total variation norm , is a real Banach space, which is isomorphic to the dual of both and its completion , thus

Remark 6Note that the previous exercise generalises the identifications from previous notes. For compact Hausdorff spaces , we have , and thus . For locally compact Hausdorff spaces that are -compact but not compact, we instead have , where is the Stone-Cech compactification of , which we will discuss in later notes.

Remark 7One can of course also define complex Radon measures to be those complex finite Borel measures whose real and imaginary parts are signed Radon measures, and define to be the space of all such measures; then one has analogues of the above identifications. We omit the details.

Exercise 18Let be two locally compact Hausdorff spaces that are also -compact, and let be a continuous map. If is an unsigned finite Radon measure on , show that the pushforward measure on , defined by , is a Radon measure on . What happens for infinite measures? Establish the same fact for signed finite Radon measures.

Let be locally compact Hausdorff and -compact. As is equivalent to the dual of the Banach space , it acquires a weak* topology (see Notes 11), known as the vague topology. A sequence of Radon measures then converges vaguely to a limit if and only if for all .

Exercise 19Let be Lebesgue measure on the real line (with the usual topology).

- Show that the measures converge vaguely as to the Dirac mass at the origin .
- Show that the measures converge vaguely as to the measure . (
Hint:Continuous, compactly supported functions are Riemann integrable.)- Show that the measures converge vaguely as to the zero measure .

Exercise 20Let be locally compact Hausdorff and -compact. Show that for every unsigned Radon measure , the map defined by sending to the measure is an isometry, thus can be identified with a subspace of . Show that this subspace is closed in the norm topology, but give an example to show that it need not be closed in the vague topology. Show that , where ranges over all unsigned Radon measures on ; thus one can think of as many ‘s “glued together”.

Exercise 21Let be a locally compact Hausdorff space which is -compact. Let be a sequence of functions, and let be another function. Show that converges weakly to in if and only if the are uniformly bounded and converge pointwise to .

Exercise 22Let be a locally compact metric space which is -compact.

- Show that the space of finitely supported measures in is a dense subset of in the vague topology.
- Show that a Radon probability measure in can be expressed as the vague limit of a sequence of discrete (i.e. finitely supported) probability measures.

** — 3. The Stone-Weierstrass theorem — **

We have already seen how rough functions (e.g. functions) can be approximated by continuous functions. Now we study in turn how continuous functions can be approximated by even more special functions, such as polynomials. The natural topology to work with here is the uniform topology (since uniform limits of continuous functions are continuous).

For non-compact spaces, such as , it is usually not possible to approximate continuous functions uniformly by a smaller class of functions. For instance, the function cannot be approximated uniformly by polynomials on , since is bounded, the only bounded polynomials are the constants, and constants cannot converge to anything other than another constant. On the other hand, on a compact domain such as , one can easily approximate uniformly by polynomials, for instance by using Taylor series. So we will focus instead on compact Hausdorff spaces such as , in which continuous functions are automatically bounded.

The space of (real-valued) polynomials is a subspace of the Banach space . But it is also closed under pointwise multiplication , making an algebra, and not merely a vector space. We can then rephrase the classical Weierstrass approximation theorem as the assertion that is dense in .

One can then ask the more general question of when a sub-algebra of – i.e. a subspace closed under pointwise multiplication – is dense. Not every sub-algebra is dense: the algebra of constants, for instance, will not be dense in when has at least two points. Another example in a similar spirit: given two distinct points in , the space is a sub-algebra of , but it is not dense, because it is already closed, and cannot *separate* and in the sense that it cannot produce a function that assigns different values to and .

The remarkable Stone-Weierstrass theorem shows that this inability to separate points is the *only* obstruction to density, at least for algebras with the identity.

Theorem 10 (Stone-Weierstrass theorem, real version)Let be a compact Hausdorff space, and let be a sub-algebra of which contains the constant function and separates points (i.e. for every distinct , there exists at least one in such that . Then is dense in .

Remark 8Observe that this theorem contains the Weierstrass approximation theorem as a special case, since the algebra of polynomials clearly separates points. Indeed, we will use (a very special case) of the Weierstrass approximation theorem in the proof.

*Proof:* It suffices to verify the claim for algebras which are closed in the topology, since the claim follows in the general case by replacing with its closure (note that the closure of an algebra is still an algebra).

Observe from the Weierstrass approximation theorem that on any bounded interval , the function can be expressed as the uniform limit of polynomials ; one can even write down explicit formulae for such a , though we will not need such formulae here. Since continuous functions on the compact space are bounded, this implies that for any , the function is the uniform limit of polynomial combinations of . As is an algebra, the lie in ; as is closed; we see that lies in .

Using the identities , , we conclude that is a *lattice* in the sense that one has whenever .

Now let and . We would like to find such that for all .

Given any two points , we can at least find a function such that and ; this follows since the vector space separates points and also contains the identity function (the case needs to be treated separately). We now use these functions to build the approximant . First, observe from continuity that for every there exists an open neighbourhood of such that for all . By compactness, for any fixed we can cover by a finite number of these . Taking the max of all the associated to this finite subcover, we create another function such that and for all . By continuity, we can find an open neighbourhood of such that for all . Again applying compactness, we can cover by a finite number of the ; taking the min of all the associated to this finite subcover we obtain with for all , and the claim follows.

There is an analogue of the Stone-Weierstrass theorem for algebras that do not contain the identity:

Exercise 23Let be a compact Hausdorff space, and let be a closed sub-algebra of which separates points but does not contain the identity. Show that there exists a unique such that .

The Stone-Weierstrass theorem is not true as stated in the complex case. For instance, the space of complex-valued functions on the closed unit disk has a closed proper sub-algebra that separates points, namely the algebra of functions in that are holomorphic on the interior of this disk. Indeed, by Cauchy’s theorem and its converse (Morera’s theorem), a function lies in if and only if for every closed contour in , and one easily verifies that this implies that is closed; meanwhile, the holomorphic function separates all points. However, the Stone-Weierstrass theorem can be recovered in the complex case by adding one further axiom, namely that the algebra be closed under conjugation:

Exercise 24 (Stone-Weierstrass theorem, complex version)Let be a compact Hausdorff space, and let be a complex sub-algebra of which contains the constant function , separates points, and is closed under the conjugation operation . Then is dense in .

Exercise 25Let be the space of trigonometric polynomials on the unit circle , where and the are complex numbers. Show that is dense in (with the uniform topology), and that is dense in (with the topology) for all .

Exercise 26Let be a locally compact Hausdorff space that is -compact, and let be a sub-algebra of which separates points and contains the identity function. Show that for every function there exists a sequence which converges to uniformly on compact subsets of .

Exercise 27Let be compact Hausdorff spaces. Show that every function can be expressed as the uniform limit of functions of the form , where and .

Exercise 28Let be a family of compact Hausdorff spaces, and let be the product space (with the product topology). Let . Show that can be expressed as the uniform limit of continuous functions , each of which only depend on finitely many of the coordinates in , thus there exists a finite subset of and a continuous function such that for all .

One useful application of the Stone-Weierstrass theorem is to demonstrate separability of spaces such as .

Proposition 11Let be a compact metric space. Then and are separable.

*Proof:* It suffices to show that is separable. By Lemma 4 of Notes 10, has a countable dense subset . By Urysohn’s lemma, for each we can find a function which equals on and is supported on . The can then easily be verified to separate points, and so by the Stone-Weierstrass theorem, the algebra of polynomial combinations of the in are dense; this implies that the algebra of *rational* polynomial combinations of the are dense, and the claim follows.

Combining this with the Riesz representation theorem and the sequential Banach-Alaoglu theorem, we obtain

Corollary 12If is a compact metric space, then the closed unit ball of is sequentially compact in the vague topology.

Combining this with Theorem 7, we conclude a special case of Prokhorov’s theorem:

Corollary 13 (Prokhorov’s theorem, compact case)Let be a compact metric space, and let be a sequence of Borel (hence Radon) probability measures on . Then there exists a subsequence of which converge vaguely to another Borel probability measure .

Exercise 29 (Prokhorov’s theorem, non-compact case)Let be a locally compact metric space which is -compact, and let be a sequence of Borel probability measures. We assume that the sequence is tight, which means that for every there exists a compact set such that for all . Show that there is a subsequence of which converges vaguely to another Borel probability measure . If tightness is not assumed, show that there is a subsequence which converges vaguely to a non-negative Borel measure , but give an example to show that this measure need not be a probability measure.

This theorem can be used to establish Helly’s selection theorem:

Exercise 30 (Helly’s selection theorem)Let be a sequence of functions whose total variation is uniformly bounded in , and which is bounded at one point (i.e. is bounded). Show that there exists a subsequence of which converges pointwise almost everywhere on compact subsets of . (Hint:one can deduce this from Prokhorov’s theorem using the fundamental theorem of calculus for functions of bounded variation.)

** — 4. The commutative Gelfand-Naimark theorem (optional) — **

One particularly beautiful application of the machinery developed in the last few notes is the commutative Gelfand-Naimark theorem, that classifies commutative -algebras, and is of importance in spectral theory, operator algebras, and quantum mechanics.

Definition 14Acomplex Banach algebrais a complex Banach space which is also a complex algebra, such that for all . An algebra isunitalif it contains a multiplicative identity , andcommutativeif for all . A -algebra is a complex Banach algebra with an anti-homomorphism map from to (thus , , and for and ) which is an isometry (thus for all ), an involution (thus for all ), and obeys theidentityfor all .A

homomorphismbetween two -algebras is a continuous algebra homomorphism such that for all . Anisomorphismis an homomorphism whose inverse exists and is also a homomorphism; two -algebras areisomorphicif there exists an isomorphism between them.

Exercise 31If is a Hilbert space, and is the algebra of bounded linear operators on this space, with the adjoint map and the operator norm, show that is a unital -algebra (not necessarily commutative). Indeed, one can think of -algebras as an abstraction of a space of bounded linear operators on a Hilbert space (this is basically the content of the non-commutative Gelfand-Naimark theorem, which we will not discuss here).

Exercise 32If is a compact Hausdorff space, show that is a unital commutative -algebra, with involution .

The remarkable (unital commutative) Gelfand-Naimark theorem asserts the converse statement to Exercise 32:

Theorem 15 (Unital commutative Gelfand-Naimark theorem)Every unital commutative -algebra is isomorphic to for some compact Hausdorff space .

There are analogues of this theorem for non-unital or non-commutative -algebras, but for simplicity we shall restrict attention to the unital commutative case. We first need some spectral theory.

Exercise 33Let be a unital Banach algebra. Show that if is such that , then is invertible. (Hint:use Neumann series.) Conclude that the space of invertible elements of is open.

Define the spectrum of an element to be the set of all such that is not invertible.

Exercise 34If is a unital Banach algebra and , show that is a compact subset of that is contained inside the disk .

Exercise 35 (Beurling-Gelfand spectral radius formula)If is a unital Banach algebra and , show that is non-empty with . (Hint:To get the upper bound, observe that if is invertible for some , then so is , then use Exercise 34. To get the lower bound, first observe that for any , the function is holomorphic on the complement of , which is already enough (with Liouville’s theorem) to show that is non-empty. Let be arbitrary, then use Laurent series to show that for all and some independent of . Then divide by and use the uniform boundedness principle to conclude.)

Exercise 36 (-algebra spectral radius formula)Let be a unital -algebra. Show thatfor all and . Conclude that any homomorphism between -algebras has operator norm at most . Also conclude that

when is self-adjoint.

The next important concept is that of a character.

Definition 16Let be a unital commutative -algebra. Acharacterof is be an element in the dual Banach space such that , , and for all ; equivalently, a character is a homomorphism from to (viewed as a (unital) algebra). We let be the space of all characters; this space is known as thespectrumof .

Exercise 37If is a unital commutative -algebra, show that is a compact Hausdorff subset of in the weak-* topology. (Hint:first use the spectral radius formula to show that all characters have operator norm , then use the Banach-Alaoglu theorem.)

Exercise 38Define an ideal of a unital commutative -algebra to be a proper subspace of such that for all and . Show that if , then the kernel is a maximal ideal in ; conversely, if is a maximal ideal in , show that is closed, and there is exactly one such that . Thus the spectrum of can be canonically identified with the space of maximal ideals in .

Exercise 39Let be a compact Hausdorff space, and let be the -algebra . Show that for each , the operation is a character of . Show that the map is a homeomorphism from to ; thus the spectrum of can be canonically identified with . (Hint:use Exercise 23 to show the surjectivity of , Urysohn’s lemma to show injectivity, and Corollary 2 of Notes 10 to show the homeomorphism property.)

Inspired by the above exercise, we define the Gelfand representation , by the formula .

Exercise 40Show that if is a unital commutative -algebra, then the Gelfand representation is a homomorphism of -algebras.

Exercise 41Let be a non-invertible element of a unital commutative -algebra . Show that vanishes at some . (Hint:the set is a proper ideal of , and thus by Zorn’s lemma is contained in a maximal ideal.)

Exercise 42Show that if is a unital commutative -algebra, then the Gelfand representation is an isometry. (Hint:use Exercise 36 and Exercise 41.)

Exercise 43Use the complex Stone-Weierstrass theorem and Exercises 40, 42 to conclude the proof of Theorem 15.

## 81 comments

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2 March, 2009 at 2:57 pm

studentIn example one, I guess it should be ”……… absolutely cts with respect to ”

[Corrected, thanks – T.]2 March, 2009 at 9:04 pm

AnonymousI gather from exercise 35 that characters are required to be homeomorphisms onto , and not just to . Is that correct?

2 March, 2009 at 10:39 pm

Terence TaoYes; I forgot to mention that characters must preserve the unit, which is equivalent to them being onto (or to them being non-zero). [I assume you meant “homomorphism” rather than “homeomorphism”.]

2 March, 2009 at 11:15 pm

Max BaroiYes, I did mean homomorphism. Sorry about that.

3 March, 2009 at 2:06 pm

mmailliw/williamI’m pretty sure R^N is metrizable (with the countable product metric), being a countable product of metric spaces; therefore, isn’t it normal (making Exercise 3 false as written)?

3 March, 2009 at 3:11 pm

Terence TaoHmm, right, the argument I had in mind was rubbish. I looked up the proof that is non-normal (which was what I was really after); it was surprisingly non-trivial. Still, I’ve outlined how it goes in the updated version of the exercises.

3 March, 2009 at 5:51 pm

EricHere is a simpler if less elementary example of a non-normal but otherwise nice (eg, locally compact Hausdorff) space. Let , where $\omega+1$ and $\omega_1+1$ are ordinals with the order topology, and . Then Y is locally compact Hausdorff since X is compact Hausdorff. However, the disjoint closed sets and (ie, the “top” and “right” edges of the product) cannot be separated by open sets. Indeed, a neighborhood of C must contain a rectangle since for each first coordinate n it contains some and there are only countably many n.

(If you prefer, you can do the same thing replacing by [0,1] and by the closed long line.)

4 March, 2009 at 9:49 pm

Anush TserunyanI have some questions about Exercise 4 (R^R is non-normal). I would really appreciate if you answer to them.

1. In the definition of K_j when you say n_x = j for all x outside of a FINITE set, shouldn’t it be n_x = j for all x outside of a COUNTABLE set instead? Otherwise, I think K_j is not closed (take an infinite countable subset A of R and define f in N^R to be an injection into N on A and j outside A, then f doesn’t belong to K_j but belongs to its closure).

2. Isn’t the third statement (the one about a sequence of disjoint finite sets A_1, A_2,…) true for all open U containing the constant 1 function?

4 March, 2009 at 10:10 pm

Terence TaoDear Anush,

Yes, the n_x should be equal to j outside of a countable set rather than a finite one. In the third statement, I forgot to add that the n_x there need to be injective on the union of the A_i’s; I’ve changed the text to reflect this.

8 March, 2009 at 2:08 am

PierreFor those who read french, you should read the very good book of laurent Schwartz “Analyse 3” Theorie de la mesure, which is a good survey of the measure theory, in every meaning (Lebesgue, Radon-Nikodym, Borel, Lusin…)

If I remember well, he uses Caratheodory’s theorem to show the Riesz representation.

18 March, 2009 at 10:32 pm

245B, Notes 13: Compactification and metrisation (optional) « What’s new[…] Show that is a unital commutative -algebra (see Section 4 of Notes 12). […]

6 April, 2009 at 2:58 pm

254C, Notes 2: The Fourier transform « What’s new[…] probability measure such that for all . (Hint: Use Prokhorov’s theorem, see Corollary 13 of 245B Notes 12. Try the case […]

12 May, 2009 at 12:01 pm

PDEbeginnerDear Prof. Tao,

I think in proof of Prop 3, should be replaced by . I still have problem on the exercise 6: We restrict our argument on a compact set (say ), according to the Urysohn’s theorem we can construct some function . Our aim is to construct a funcition in $C_c(X)$, it seems we can set for . I am a little doubted if this new is continuous (It most probably a trivial question).

12 May, 2009 at 12:10 pm

Terence TaoThanks for the correction! To make f continuous across all of X, one needs to ensure that f vanishes on the boundary of A, but Urysohn’s lemma can guarantee this for us.

24 May, 2009 at 1:55 am

PDEbeginnerDear Prof. Tao,

I was wondering if you could give a little more details on the hint of Exercise 8.

I also have some problem on exercise 9: First extend the to be a new function on by Tietze extension theorem, there may be for some . Define , which is a closed set. By Urysohn’s lemma, we can construct a continuous funtion such that for and for , where is any closed subset of . Let , and let , but we have the possibility for some by the construction of . Most probably my understanding your hint is not right.

Thanks a lot in advance!

Best regards.

24 May, 2009 at 9:12 am

Terence TaoFor Exercise 8, you might want to play with some simple examples, e.g. when X consists of K together with a single additional point x, in which case the only task is to figure out what value to assign to . Try to make as large as possible while retaining the Lipschitz property. Once you have the formula for in this one-point-extension case, you should be able to handle the general case.

For Exercise 9, set B equal to K.

24 May, 2009 at 10:28 pm

PDEbeginnerDear Prof. Tao,

As for Exercise 9, I am doubted if there are some point and some sequence with such that (or ). In this case, there seems some problem for taking as .

24 May, 2009 at 11:37 pm

Terence TaoDear PDEbeginner,

By hypothesis, K is closed, and f is continuous and finite on K, so the scenario you describe cannot occur.

19 June, 2009 at 5:47 am

Haokun XuHi Professor Tao,

In your proof of the Stone-Weierstrass theorem, I think you meant to define min(f,g) = (f+g)/2 – |(f-g)/2|, as opposed to (f+g)/2 + |(f-g)/2|.

19 June, 2009 at 5:58 am

Haokun XuHi Professor Tao,

Also, in the last paragraph of your proof of the Stone-Weierstrass theorem, with the sentence beginning “First, observe from continuity that,” I think you meant g_xy (y’) > f(y’) + epsilon as opposed to g(y’) > f(y’) + epsilon.

19 June, 2009 at 7:17 am

Terence TaoDear Haokun: thanks for the corrections!

19 June, 2009 at 5:13 pm

Haokun XuHi Professor Tao,

Also, in the first paragraph of your proof of Urysohn’s lemma, I think you meant to say that (3)(4) and (1)(2) are clearly equivalent as oppose to (2)(3) and (1)(2).

[Corrected, thanks – T.]4 August, 2009 at 2:27 pm

PDEbeginnerDear Prof. Tao,

Could you give some hint on the proof of exercise 23? In section 3 (paragraph 4), it seems that the sub-algebra is the subspace of closed under pointwise multiplication and addition.

Thanks.

4 August, 2009 at 4:02 pm

Terence TaoThere was a slight error in that exercise: I forgot to add the assumption that had to be closed.

One can establish the claim in two parts. First, show that is contained in one of the by arguing by contradiction (if this was not the case, one could soon create a function in the algebra that never vanished, and then show that the algebra contained 1, a contradiction). Then, adjoin the constants to the algebra and use the previous Stone-Weierstrass theorem (Theorem 10).

4 August, 2009 at 9:46 pm

PDEbeginnerDear Prof. Tao,

I tried to prove the claim in a similar way before I asked you the question.

To construct a never vanishing funciton, I tried to use the argument in the proof of Theorem 10: prove that if , and then construct a never vanishing function. However, to prove if $f, g \in \mathcal A$, we seem to need the constant in $\mathcal A$.

Suppose that we have constructed a never vanishing function , I tried to apply the Taylor’s expansion for , in Taylor’s expansion of , the first term is a constant, i.e. if , then the constant needs to be in .

I will think of it more.

7 October, 2009 at 8:57 pm

Israel Gelfand « What’s new[…] Gelfand’s beautiful theory of -algebras and related spaces made quite an impact on me as a graduate student in Princeton, to the point where I was seriously considering working in this area; but there was not much activity in operator algebras at the time there, and I ended up working in harmonic analysis under Eli Stein instead. (Though I am currently involved in another operator algebras project, of which I hope to be able to discuss in the near future. The commutative version of this theory is discussed in these lecture notes of mine.) […]

20 January, 2010 at 8:06 am

Seminar „Funktionalanalysis“ « UGroh's Weblog[…] auf kompakten Hausdorffräumen (T. Tao, Continuous functions on locally compact Hausdorff spaces, Blognotes zur Vorlesung 245B; N. Carothers, l.c. Chap. 12 […]

25 June, 2010 at 6:44 pm

The uncertainty principle « What’s new[…] Stone/Gelfand duality A (locally compact Hausdorff) topological space can be viewed in physical space (as a collection of points), or dually, via the algebra of continuous complex-valued functions on that space, or (in the case when is compact and totally disconnected) via the boolean algebra of clopen sets (or equivalently, the idempotents of ). The fundamental connection between the two is given by the Stone representation theorem or the (commutative) Gelfand-Naimark theorem. […]

24 August, 2010 at 3:59 am

Kestutis CesnaviciusThank you for a wonderful set of notes! A couple of (possibly) typos:

1. In Exercise 7 should be .

2. In Lemma 6 I don’t see how the given conditions imply continuity of . I guess each should have a neighbourhood intersecting only finitely many ‘s.

3. In the last display in the proof of Theorem 7 the middle union should be to rather than to .

4. Towards the end of the proof of Theorem 8 in two instances the indices are missing. Also, once should be in the same (penultimate) paragraph.

5. In Remark 5 should be .

6. In the second sentence of the proof of Lemma 9 and should be swapped in the inequality.

7. Exercise 18 does not hold as stated (e.g., maps the plane to the origin).

8. In Exercise 25 one should impose additionally or replace the unit interval by the circle.

9. The Corollary 12, I guess, should be “If is a compact metric space, then the closed unit ball in is sequentially compact in the vague topology.”

10. Exercise 30 does not hold as stated: e.g., a family of piecewise linear functions, the nth of which passes through (-2/n, 0), (-1/n, 1), (1/n, -1), (2/n, 0) and is linear inbetween (and zero at infinities).

11. It seems that (xy)* = y*x* is missing in the Definition 14.

12. In Exercise 35 should be .

[Corrected, thanks – T.]9 September, 2010 at 10:54 pm

245A, Notes 1: Lebesgue measure « What’s new[…] of measure can be used to define the concept of a Radon measure, which will be encountered much later in this course series. Exercise 14 (Criteria for finite measure) Let . Show that the following are […]

19 September, 2010 at 7:22 pm

245A, Notes 2: The Lebesgue integral « What’s new[…] But then, by Exercise 14 of http://en.wikipedia.org/wiki/Urysohn’s_lemma, which we wil cover in 245B). It is then clear from construction that as […]

25 September, 2010 at 10:58 pm

245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new[…] One question that will not be addressed much in this current set of notes is how one actually constructs interesting examples of measures. We will discuss this issue more in later notes (although one of the most powerful tools for such constructions, namely the Riesz representation theorem, will not be covered until 245B). […]

10 October, 2010 at 9:45 am

yaoliangI was trying to do Exercise 11. Do we need to assume the l.s.c. function f is bounded from below? Otherwise I can’t figure out how to avoid function g to take value -infty. As a side note, the condition on X can be relaxed to completely regular (which turns out to be necessary as well).

Thanks.

10 October, 2010 at 9:58 am

Terence TaoAh, yes, one does need to allow g to take the value -infty in this case. I’ve adjusted the question accordingly.

16 October, 2010 at 8:29 pm

245A, Notes 5: Differentiation theorems « What’s new[…] finite Borel measures have this property, but we will not prove this here; see Exercise 12 of these notes.) Establish the Hardy-Littlewood maximal […]

30 October, 2010 at 6:54 pm

245A, Notes 6: Outer measures, pre-measures, and product measures « What’s new[…] We remark that these notes omit a very important way to construct measures, namely the Riesz representation theorem, but we will defer discussion of this theorem to 245B. […]

16 December, 2010 at 4:30 am

245A, Notes 2: The Lebesgue integral « mathTHÍCHinTOÁNmyHỌCbrain[…] can set for some sufficiently large ; one may also invoke Urysohn’s lemma, which we wil cover in 245B). It is then clear from construction that as […]

2 April, 2011 at 6:39 pm

Peter LuHi prof Tao,

I’m not sure if you’ve been informed yet, but I believe there is a misnumbering in your homework for 245C. The last problem given was “Exercise 1.10.25 (Density of Fourier series). Note an important typo in this question: [R,Z] should be R/Z (i.e. the unit circle).” which is clearly Exercise 25 in the notes of this blog post. It seems to be a different problem in your book, or at least the pdf copy that you have available for download.

2 April, 2011 at 6:43 pm

Peter LuOh my mistake, I went back and checked again and it is more or less the same problem except you have . C[0,1] instead of R/Z in the book.

7 June, 2011 at 7:33 am

Central extensions of Lie groups, and cocycle averaging « What’s new[…] Thus, we may then apply the (LCH version of the) Tietze extension theorem (discussed on this blog here), and extend the map to a continuous (and compactly supported) function defined on all of […]

16 July, 2011 at 7:20 am

AnonymousHello,

In the proof of the existence of the partitions of unity, by “[…] is well-defined, continuous and bounded below by 1” did you mean that is just bounded? a point may belong to multiple compact sets (but to only a finite amount of them).

[No, it is important that is both continuous and bounded below in order to have a continuous inverse. It is not necessarily bounded above, but it is finite at every point. -T]23 July, 2011 at 6:41 am

AnonymousHello, I have an alternative proof of the subaditivity of the linear functional from the RRT, I think it is ok, here it is:

Lemma:If are such that , then .Proof:If then equality is trivialy true, so we can assume and the series converges.Because of the way is defined, it es enough to proof that if is such that then there exist such that and .

Consider the family

which is not empty because the sequence of null functions is inside

. is partially ordered by if for all . Let

a totally ordered subset of .Then

is a nondecreasing sequence of continuos functions on a compact set, so by Dini’s theorem in converged uniformly to a continuous function , and clearly . I claim that , indeed, if

for some and for all sufficiently small \epsilon greater than 0, then as every term involved is nonnegative and for all , there exists some such that

. By the lower semicontinuity of there is an open neighbourhood of such that for every , and by the continuity of ,

changing if necessary, we have that for any ,

.

As locally compact, there is a compact neighbourhood of

in , and by normality of and Urysohn's lemma there is a function such that

and is zero in the complement of .Taking one gets an element in

strictly greater than , which proves that

as we wished. []

16 July, 2015 at 10:44 pm

AnonI don’t think your application of Dini’s theorem is correct, continuity of the limit is an assumption of Dini’s theorem and not part of the conclusion (uniform convergence is the only part of the conclusion of Dini’s theorem).

E.g: Take the function $1-x^n$ on the closed unit interval, it is an increasing sequence of functions but the limit is not continuous.

9 September, 2011 at 4:05 am

Georges ElencwajgAlgebraically minded readers might like to notice that, given an arbitrary affine scheme X=Spec(A) and two disjoint closed subschemes Y=V(I), Z =V(J), there exists a global regular function f in A whose restrictions to Y and Z are respectively 0 and 1. This follows from the Chinese remainder theorem since disjointness of Y and Z translates into I+J=A .

Of course if X is irreducible, it is impossible to find disjoint neighbourhoods of Y and Z , since there are no disjoint nonempty open sets in X at all.

More generally the analogon of Tietze’s theorem is true: any regular function on Y extends to a regular function on X. (Affine) scheme theory is built in such a way that this is actually a tautology , boiling down to the fact that the quotient morphism from A to A/I is surjective!

11 February, 2012 at 1:46 am

RexIn the proof of Theorem 7, you remark “(note that as {X} is compact, a subset {K} of {X} is closed if and only if it is compact).”

You probably want to mention here that you are also using the fact that X is Hausdorff to get the implication “K compact ==> K closed”. In general, a compact subset K of a compact space X will not be closed.

[Corrected, thanks – T.]11 February, 2012 at 2:56 am

RexIn Exercise 17, you write

“{C_c(X \rightarrow {\bf R})} and its completion {C_0(X \rightarrow {\bf R})}”

What norm on C_c(X –> R) are we taking its completion with respect to?

[ is understood to be given the uniform topology by default. -T]13 February, 2012 at 3:42 am

RexI think that Proposition 3 could be simplified a bit. I’ll reproduce it, as written above, here for convenience:

“Proposition 3: Let be a locally compact Hausdorff space which is also -compact, and let be a Radon measure on . Then for any , is a dense subset in (real-valued) . In other words, every element of can be expressed as a limit (in ) of continuous functions of compact support.

Proof: Since continuous functions of compact support are bounded, and compact sets have finite measure, we see that is a subspace of . We need to show that the closure of this space contains all of .

Let be a compact set, and let be a Borel set, then has finite measure. Applying inner and outer regularity, we can find a sequence of compact sets and open sets such that . Applying Exercise 6, we can then find such that . In particular, this implies (by the squeeze theorem) that converges in to (here we use the finiteness of ); thus lies in for any measurable subset of . By linearity, all simple functions supported on also lie in ; taking closures, we see that any function supported in also lies in . As is -finite, one can express any non-negative function as a monotone limit of compactly supported functions, and thus every non-negative function lies in , and thus all functions lie in this space, and the claim follows. "

It seems that the compact set here is extraneous; one only needs the fact that has finite measure to apply the argument about approximation via inner/outer regularity and bump functions. I don't see how plays any role.

If this is correct, then we should have that lies in the closure for all sets of finite measure. Consequently, the space of simple functions with finite measure support lies in . But you have already proven in Proposition 2 of

https://terrytao.wordpress.com/2009/01/09/245b-notes-3-lp-spaces/

that the closure of the space of simple functions with finite measure support in is, in fact, all of (provided that ). The proof you give there works for an arbitrary measure space.

In particular, we can avoid the -compact hypothesis entirely.

[Fair enough; I’ve adjusted the text accordingly. -T.]26 June, 2012 at 8:49 pm

Qiaochu YuanRegarding Exercise 37, Banach-Alaoglu is unnecessary. The weak-* topology on the space of characters is just the Zariski topology on the maximal spectrum, and this is compact for any commutative ring whatsoever by a straightforward and purely algebraic argument.

26 June, 2012 at 10:08 pm

Terence TaoWell, OK, although showing that the weak-* topology and Zariski topology are equivalent on is not entirely trivial (I think one needs to at least develop a continuous functional calculus for this).

27 June, 2012 at 11:15 am

Qiaochu YuanAh. My apologies. I was under the mistaken impression that the two were equivalent by definition (but the weak-* topology is the coarsest topology making evaluation on elements of continuous when is given the usual topology and the Zariski topology is the coarsest topology making evaluation on elements of continuous when is given the

Zariskitopology…).17 July, 2012 at 2:25 pm

SvenDear Prof. Tao,

Helly’s selection theorem seems to always be stated for functions on an interval. Does that mean that it cannot be generalized to functions on $\mathbb{R}^N$?

17 July, 2012 at 2:41 pm

Terence TaoIn higher dimensions, the total variation norm does not control the uniform norm (the Sobolev embedding numerology becomes increasingly unfavorable as the dimension increases), and so a naive extension of Helly’s theorem to higher dimensions fails. But there are certainly other compactness theorems in higher dimensions which can be viewed as substitutes (e.g. Rellich compactness theorem).

18 August, 2012 at 9:07 pm

Noncommutative probability « Annoying Precision[…] In fact a much stronger statement is true due to the following corollary of the Riesz-Markov theorem, which we will not prove; see Terence Tao’s notes. […]

28 October, 2012 at 4:53 am

KatyDear Prof. Tao,

Thank you for the wonderful notes! In exercise 36, I think that $||x|| = \{ \sup |z| : z \in \sigma(x) \}$ holds for self-adjoint elements of the C*-algebra. For example, consider the operator $x$ on $\mathbb{C}^2$ that sends $(e_1,e_2) \to (e_2, 0)$. In this case, $\sigma(x) = 0$, but $||x|| = 1$.

[Corrected, thanks – T.]18 January, 2013 at 7:54 am

LugDear Dr. Tao,

This might just be my problem, but In exercise 3, I found the word ‘refinement’ confusing, which would make people (at least me) think F’ is coarser than F, but actually F’ is finer. I actually struggled a while there for a while by the contradicting meanings of the two definitions of F’.

23 April, 2013 at 5:26 pm

Solution Set I | Interesting Mathematics exercises[…] : Initial hunch is to hammer this problem to with the Weierstrass Approximation Theorem (see Terrence Tao’s blog for a more general version of the same). However, one needs to be careful in getting uniformity in approximation. Any simplification of […]

18 July, 2013 at 3:36 am

Anonymoussir for a function belongs to the test space (that is space of all continuous function with compacct support) i want to decompose one function say “\phi” into difference of two non-negative continuous function with compact support.what is this positive and negative part of function \phi) ?

29 August, 2013 at 11:58 am

UPSDear Prof. Tao,

Does there exists an extension of the Prokhorov’s theorem to the space of finitely additive signed measures (with an underlying space say R^n)?

Motivation: Bounded Lipschitz (dual) metric metrizes the narrow topology on the space of finitely additive signed measures. So can we have a similar characterization (like Prokhorov’s theorem) of compact sets in this space.

23 September, 2014 at 1:53 pm

Richard HevenerIn your proof of Thm. 5 (Lusin), line 5, you should change ‘By Theorem 4’ to ‘By Exercise 10’. Also, why not drop the -compact hypothesis?

[Corrected, thanks. For some minor technical reasons I have restricted Radon measures here to the sigma-compact case; see Example 1 and Remark 3. -T.]1 January, 2015 at 8:08 am

AlexIn Lemma 9, you don’t really need to equip with a topology in order to show that every linear functional can be written as a difference of two positive functionals and . If is equipped with the uniform norm then is continuous if and only if the corresponding Radon measure is a finite signed measure. However, we can also equip with a locally convex topology (see http://en.wikipedia.org/wiki/Radon_measure#Measures). In this case, we have more continuous functionals, namely signed Radon measures that can be also infinite on one side or even “real” Radon measures, i.e. differences of positive Radon measures, that cannot be written in terms of a set function.

23 April, 2015 at 5:22 am

BobDear Prof. Tao,

can you give me a reference about the proof of Riesz representation theorem via creating L^1 as the completion of C_c(X)?

Thanks

24 April, 2015 at 9:21 pm

eshetuCould anyone help me the proof of theorem 15? Thanks.

12 October, 2015 at 12:35 pm

275A, Notes 2: Product measures and independence | What's new[…] for any compact separable metric spaces. Radon measures are often used in real analysis (see e.g. these lecture notes) but we will not develop their theory further […]

2 November, 2015 at 7:06 pm

275A, Notes 4: The central limit theorem | What's new[…] distribution (this is a consequence of the Riesz representation theorem, discussed for instance in this blog post). As a consequence of the insensitivity of convergence in distribution to equivalence in […]

13 November, 2015 at 9:06 am

AnonymousThe formulation of partitions of unity looks quite different from the ones from the Wikipedia link you gave.

What is essential with using the “closed” cover in Lemma 6? Can one formulate it by using only “open” cover?

I’ve seen the

smooth partitions of unityin PDE (for example in Evans’s book) which says that if is an open cover of an open subset of , then there exists such that and the sum of equals to pointwise on $U$. Can this be deduced from Lemma 6 (and suitable mollification)?13 November, 2015 at 9:59 am

Terence TaoIf the space is locally compact Hausdorff, then any open cover will have a partition of unity with compactly supported functions, and the additional conditions in Lemma 6 are not needed. (Actually, this would make a good additional exercise here, which I may add to a future edition of the text.) But in the normal case, I believe some condition like this is necessary.

One can obtain a smooth partition of unity from a continuous one by convolving all the functions by an smooth approximation to the identity (with the exact choice of mollifier dependent on the location, as the open sets may become increasingly small in diameter at infinity), and then dividing out by the sum as before in order to recover the property of summing to one.

2 December, 2015 at 7:08 am

AnonymousIn the proof of the Urysohn lemma, can

be replaced with

or

?

[Yes, although these descriptions of are not as useful as the first two for establishing continuity. -T.]25 February, 2016 at 4:37 pm

AnonymousDo you have a reference for part 4 (The commutative Gelfand-Naimark theorem (optional) ) in this note which is also similar to the way you represent it?

26 February, 2016 at 9:38 am

Terence TaoI adapted this material from various texts on non-commutative Gelfand theory (of which there are several), specialising to the commutative case. I don’t know of a text that focuses primarily on the commutative case though.

25 February, 2016 at 5:43 pm

AnonymousDoes the “spectrum” in Definition 16 have anything to do with the one in https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Definition

?

26 February, 2016 at 9:33 am

Terence TaoThis is the Gelfand spectrum: https://en.wikipedia.org/wiki/Gelfand_representation#The_spectrum_of_a_commutative_C.2A-algebra of a function space, rather than the spectrum of a linear operator. In some cases there is a relationship; for instance the Gelfand spectrum of the convolution algebra is naturally associated to the joint spectrum of the infinitesimal translation operators . In particular, both concepts can be viewed as capturing a notion of the available “frequencies” in a function space or linear operator. However in general there is not much of a direct relationship between the two.

28 February, 2016 at 3:35 pm

AnonymousIn http://math.oregonstate.edu/~holladwi/banach_algebras.pdf , the Gelfand-Naimark Theorem (Theorem 50) is called as the “abstract spectral theorem”, in which the spectrum is the set of characters. Then Gelfand-Naimark is used in the proof of Theorem 51 (spectral theorem), in which the spectrum of a linear operator appears.

Is this also related to what you said in the comment?

25 February, 2016 at 7:32 pm

AnonymousDo Definition 16 and Exercise 39 remain the same if one replaces -algebra with Banach-algebra and drops the condition ?

26 February, 2016 at 9:37 am

Terence TaoI believe one can prove something of this form with the aid of the Riesz representation theorem, but I haven’t checked the details.

28 June, 2016 at 12:04 am

NickExercise 11 does not need to hold for non-Hausdorff . A counterexample being the two point space with one point being open but not closed. All continuous functions are constant, but there are nonconstant semicontinuos ones.

29 June, 2016 at 6:54 am

Terence TaoSorry, could you elaborate on why this is a counterexample to either part of Exercise 11? Note that part 2 of this exercise has the additional assumption that is normal.

29 June, 2016 at 9:41 am

NickIt’s a little nitpicky, but by the definition given in these notes, this space is normal (while not fulfilling part 2). It is just required that you can separate disjoint closed sets by disjoint open neighborhoods, which is automatically the case here, since one of the closed sets must be empty.

[Ah, I see now. Correction added, thanks – T.]12 November, 2016 at 4:44 am

zhcheDear Prof Tao,

As for exercise 15, the uniqueness of the minimal decomposition is quite easy to prove. But for the existence part, I think for any two different decompositions and , they only need to satisfy that difference is a linear functional, not necessarily the value be positive for each in . Then the sense of ordering does not necessarily for any two decompositions, i.e., there may be no minimal decompositions. Am I understanding right ?

Regards

Che Zhihua

[It is possible for a partial order to have a minimal element even if it fails to be a total order. For instance, the set of all subsets of a given set (partially ordered by set inclusion) has a minimal element even though most pairs of subsets of are incomparable. -T.]13 November, 2016 at 5:06 am

zhcheYes, I found it partial order. So this exercise means to want us to prove that this poset HAS a least element, e.g., prove it is bounded lattice ? Then this exercise is not that trivial.

4 December, 2016 at 12:07 am

IanProfessor Tao:

Concerning the proof of Lusin’s Theorem, could you please explain the use of Exercise 10? It seems to me that this exercise produces a function that agrees with only in (most of) its compact support, but doesn’t really guarantee anything on the whole space.

4 December, 2016 at 1:22 pm

Terence TaoOne can multiply the function produced by Exercise 10 with a cutoff function coming from Urysohn’s lemma to make the support of that function only slightly larger than the support of .