It occurred to me recently that the mathematical blog medium may be a good venue not just for expository “short stories” on mathematical concepts or results, but also for more technical discussions of individual mathematical “tricks”, which would otherwise not be significant enough to warrant a publication-length (and publication-quality) article. So I thought today that I would discuss the *amplification trick* in harmonic analysis and combinatorics (and in particular, in the study of estimates); this trick takes an established estimate involving an arbitrary object (such as a function f), and obtains a stronger (or *amplified*) estimate by transforming the object in a well-chosen manner (often involving some new parameters) into a new object, applying the estimate to that new object, and seeing what that estimate says about the original object (after optimising the parameters or taking a limit). The amplification trick works particularly well for estimates which enjoy some sort of symmetry on one side of the estimate that is not represented on the other side; indeed, it can be viewed as a way to “arbitrage” differing amounts of symmetry between the left- and right-hand sides of an estimate. It can also be used in the contrapositive, amplifying a weak counterexample to an estimate into a strong counterexample. This trick also sheds some light as to why dimensional analysis works; an estimate which is not dimensionally consistent can often be amplified into a stronger estimate which is dimensionally consistent; in many cases, this new estimate is so strong that it cannot in fact be true, and thus dimensionally inconsistent inequalities tend to be either false or inefficient, which is why we rarely see them. (More generally, any inequality on which a group acts on either the left or right-hand side can often be “decomposed” into the “isotypic components” of the group action, either by the amplification trick or by other related tools, such as Fourier analysis.)

The amplification trick is a deceptively simple one, but it can become particularly powerful when one is arbitraging an unintuitive symmetry, such as symmetry under tensor powers. Indeed, the “tensor power trick”, which can eliminate constants and even logarithms in an almost magical manner, can lead to some interesting proofs of sharp inequalities, which are difficult to establish by more direct means.

The most familiar example of the amplification trick in action is probably the textbook proof of the Cauchy-Schwarz inequality

(1)

for vectors v, w in a complex Hilbert space. To prove this inequality, one might start by exploiting the obvious inequality

(2)

but after expanding everything out, one only gets the weaker inequality

. (3)

Now (3) is weaker than (1) for two reasons; the left-hand side is smaller, and the right-hand side is larger (thanks to the arithmetic mean-geometric mean inequality). However, we can amplify (3) by arbitraging some symmetry imbalances. Firstly, observe that the phase rotation symmetry preserves the RHS of (3) but not the LHS. We exploit this by replacing v by in (3) for some phase to be chosen later, to obtain

.

Now we are free to choose at will (as long as it is real, of course), so it is natural to choose to optimise the inequality, which in this case means to make the left-hand side as large as possible. This is achieved by choosing to cancel the phase of , and we obtain

(4)

This is closer to (1); we have fixed the left-hand side, but the right-hand side is still too weak. But we can amplify further, by exploiting an imbalance in a different symmetry, namely the homogenisation symmetry for a scalar , which preserves the left-hand side but not the right. Inserting this transform into (4) we conclude that

where is at our disposal to choose. We can optimise in by minimising the right-hand side, and indeed one easily sees that the minimum (or infimum, if one of v and w vanishes) is (which is achieved when when are non-zero, or in an asymptotic limit or in the degenerate cases), and so we have amplified our way to the Cauchy-Schwarz inequality (1). [See also this discussion by Tim Gowers on the Cauchy-Schwarz inequality.]

— Amplification via phase, homogeneity, or dilation symmetry —

Many similar examples of amplification are used routinely to prove the basic inequalities in harmonic analysis. For instance to deduce the complex-valued triangle inequality

(5)

(where is a measure space and f is absolutely integrable) from its real-valued counterpart, we first apply the latter inequality to to obtain

.

To make the right-hand side phase-rotation-invariant, we crudely bound by , obtaining

and then one can arbitrage the imbalance in phase rotation symmetry to obtain (5). For another well-known example, to prove Hölder’s inequality

(6)

for non-negative measurable and dual exponents , one can begin with the elementary (weighted) arithmetic mean-geometric mean inequality

(7)

for non-negative a,b (which follows from the convexity of the function , which in turn follows from the convexity of the exponential function) to obtain the inequality

.

This inequality is weaker than (6) (because of (7)); but if one amplifies by arbitraging the imbalance in the homogenisation symmetry one obtains (6). As a third example, the Sobolev embedding inequality

(8)

which is valid for (and also valid in some endpoint cases) and all test functions (say) f on , can be amplified to obtain the *Gagliardo-Nirenberg inequality*

(9)

where is the number such that , by arbitraging the action of the dilation group . (In this case, the dilation action does not leave either the LHS or RHS of (8) invariant, but it affects the LHS in a well controlled manner, which can be normalised out by dividing by a suitable power of .) The same trick, incidentally, reveals why the Sobolev embedding inequality fails when or when , because in these cases it leads to an absurd version of the Gagliardo-Nirenberg inequality. Observe also that the Gagliardo-Nirenberg inequality (9) is dimensionally consistent; the dilation action affects both sides of the inequality in the same way. (The weight of the representation of the dilation action on an expression is the same thing as the exponent of the length unit that one assigns to the dimension of that expression.) More generally, arbitraging a dilation symmetry allows a dimensionally consistent inequality to emerge from a dimensionally inconsistent (or dimensionally inefficient) one.

— Amplification using linearity —

Another powerful source of amplification is *linearity* (the principle of superposition). A simple example of this is *depolarisation*. Suppose one has a symmetric bilinear form from a normed vector space X to the real numbers, and one has already proven the polarised inequality

for all f in X. One can amplify this by replacing f with f+cg for arbitrary f, g in X and a real parameter c, obtaining

Optimising this in c (e.g. taking ) and using the triangle inequality, one eventually obtains the amplified (depolarised) inequality

for some absolute constant C.

For a slightly more sophisticated example, suppose for instance that one has a linear operator for some and some measure spaces X,Y, and that one has established a scalar estimate of the form

(10)

for arbitrary scalar functions f. Then by replacing by a signed sum , where are arbitrary functions in and are signs, and using linearity, we obtain

If we raise this to the power, take the to be random (Bernoulli) signs (in order to avoid unexpectedly large cancellations in the series), and then take expectations of both sides, we obtain

If one then uses Khintchine’s inequality to compute the expectations, one ends up with the vector valued estimate

for some constant depending only on p (in particular, it is independent of N). We can then use the monotone convergence theorem to amplify the finite sum to an infinite sum, thus

.

Comparing this to (10) we see that we have amplified a scalar inequality (in which the unknown function f takes values in the real or complex numbers) to a vector-valued inequality (in which we have a sequence taking values in the Hilbert space ). [This particular amplification was first observed by Marcinkiewicz and Zygmund.]

If the estimate one is studying involves “localised” operators or norms, then one can use linearity to amplify a global estimate into a more localised one. For instance, let us return to the Sobolev inequality (8). We can establish a partition of unity for some bump function , then we see that

Applying the Sobolev inequality (8) to each localised function and then summing up, one obtains the localised Sobolev inequality

where is the cube of sidelength 1 centred at n. This estimate is a little stronger than (8), because the summation norm is smaller than the summation norm.

— Amplification via translation invariance —

If is a translation-invariant operator on which is not identically zero, one can automatically rule out a large variety of estimates concerning T due to their incompatibility with translation invariance (they would amplify themselves into an absurd estimate). For instance, it will not be possible to establish any weighted estimate involving power weights such as in which there is a higher exponent on the left. More precisely if are real numbers and , then it is not possible for any estimate of the form

to be true. Indeed, if such an estimate was true, then by using the translation invariance we can amplify the above estimate to

for any . But if one fixes f and lets go to infinity, we see that the right-hand side grows like while the left-hand side grows like (unless Tf vanishes entirely), leading to a contradiction.

[There is a Fourier dual to the above principle, familiar to experts in the analysis of PDEs, which asserts that a function space norm with a low number of derivatives (i.e. a low-regularity norm) cannot control a norm with a high number of derivatives. Here, the underlying symmetry that drives this principle is modulation invariance rather than translation invariance.]

One can obtain particularly powerful amplifications by combining translation-invariance with linearity, because one can now consider not just translates of a single function f, but also consider superpositions of such functions. For instance, we have the principle (which I believe was first articulated by Littlewood) that a non-trivial translation-invariant linear operator T can only map to when . (Littlewood summarised this principle as “the higher exponents are always on the left”.) To see this, suppose that we had an estimate of the form

(11)

We can amplify this estimate by replacing by

, where N is some integer and are widely separated points. If these points are sufficiently far apart, then the RHS of (11) is comparable to , whereas the LHS is comparable to

(note how this uses both the translation-invariance and linearity of T). Thus in the limit we obtain

Letting N go to infinity, we obtain a contradiction unless (or unless T is identically zero).

The combination of translation invariance and linearity is so strong that it can amplify even a very qualitative estimate into a quantitative one. A good example of this is Stein’s maximal principle. Suppose we have some maximal operator on some compact group G with normalised Haar measure dm, where the are a sequence of translation-invariant operators which are uniformly bounded on some space for some . Suppose we are given the very weak information that is finite almost everywhere for every . (This is for instance the case if we know that converge pointwise almost everywhere.) Miraculously, this qualitative hypothesis can be amplified into a much stronger quantitative one, namely that is of weak type :

(12)

To see this, suppose for contradiction that (12) failed for any C; by homogeneity, it would also fail even when restricted to the case . What this means is that for any , there exists such that

(13)

where is the set where .

At present, could be a very small subset of G, although we know that it has positive measure. But we can amplify this set to be very large by the following trick: pick an integer N comparable to , select N random shifts and random signs and replace by the randomised sum . This sum will tend to be large (greater than or comparable to 1) on most of the union ; this can be made precise using Khintchine’s inequality. On the other hand, another application of Khintchine’s inequality using (13) shows that has an norm of on the average. Thus we have constructed functions f of arbitrarily small norm whose maximal function Mf is bounded away from zero on a set of measure bounded away from zero. From this and some minor additional tricks it is not difficult to then construct a function f in whose maximal function is infinite on a set of positive measure, leading to the desired contradiction.

— The tensor power trick —

We now turn to a particularly cute source of amplification, namely the tensor power operation which takes a complex-valued function on some set X and replaces it with a tensor power defined by

If one has an estimate for which only one of the sides behaves nicely under tensor powers, then there can be some opportunity for arbitrage. For instance, suppose we wanted to prove the Hausdorff-Young inequality

(14)

on arbitrary finite additive groups G and all , where is the dual exponent of p, is the Pontryagin dual of G (i.e. the group of characters on G), we give G normalised counting measure, and is the Fourier transform on G. If we had the Riesz-Thorin interpolation theorem, we could quickly deduce (14) from the trivial inequality

(15)

and the Plancherel identity

(16);

indeed, this is one of the textbook applications of that theorem. But suppose for some reason one did not wish to use the Riesz-Thorin theorem (perhaps in a desire to avoid “non-elementary” methods, such as complex analysis), and instead wished to use the more elementary Marcinkiewicz interpolation theorem. Then, at first glance, it appears that one can only conclude the weaker estimate

for some constant . However, we can exploit the fact that the Fourier transform commutes with tensor powers. Indeed, by applying the above inequality with f replaced by (and G replaced by ) we see that

for every ; taking roots and then letting M go to infinity we obtain (14); the tensor power trick has “magically” deleted the constant from the inequality. More generally, one can use the tensor power trick to *deduce* the Riesz-Thorin interpolation theorem from the Marcinkiewicz interpolation theorem (the key point being that the operator norm of a tensor power of a linear operator T is just the power of the operator norm of the original operator T). This gives a proof of the Riesz-Thorin theorem that does not require complex analysis.

Actually, the tensor power trick does not just make constants disappear; it can also get rid of logarithms. Because of this, we can make the above argument even more elementary by using a very crude form of the Marcinkiewicz interpolation argument. Indeed, suppose that f is a quasi-step function, or more precisely that it is supported on some set E in G and takes values between A and 2A for some . Then from (15) and (16) we see that and , and hence . Now if f is not a quasi-step function, one can decompose it into such functions by the “wedding cake” decomposition (dividing the range of |f| into dyadic intervals from to ; the portion of |f| which is less than can be easily dealt with by crude methods). From the triangle inequality we then conclude the weak Hausdorff-Young inequality

.

If one runs the tensor power trick again, one can eliminate both the constant factor and the logarithmic factor and recover (14) (basically because converges to 1 as M goes to infinity). [More generally, the tensor power trick can convert restricted or weak-type estimates into strong-type estimates.]

The deletion of the constant may seem minor, but there are some things one can do with a sharp estimate that one cannot with a non-sharp one. For instance, by differentiating (14) at p=2 (where equality holds) one can obtain the *entropy uncertainty principle*

whenever we have the normalisation . (More generally, estimates involving Shannon entropy tend to be rather amenable to the tensor power trick.)

[I should remark that in Euclidean space, the constant in Hausdorff-Young can be improved to below 1, but this requires some particularly Euclidean devices, such as the use of Gaussians, although this is not too dissimilar as there are certainly many connections between Gaussians and tensor products (cf. the central limit theorem). All of the above discussion also has an analogue for Young’s inequality.]

The tensor power trick also allows one to *disprove *certain estimates. Observe that if two functions f, g on a finite additive group G such that for all x (i.e. g *majorises* f), then from Plancherel’s identity we have

and more generally (by using the fact that the Fourier transform intertwines convolution and multiplication) that

for all even integers . Hardy and Littlewood conjectured that a similar bound held for all , thus

But if such a bound held, then by the tensor power trick one could delete the constant . But then a direct computation (for instance, inspecting what happens when f is infinitesimally close to g) shows that this amplified estimate fails, and so the Hardy-Littlewood majorant conjecture is false. (With a little more work, one can then transfer this failure from finite abelian groups G to other groups, such as the unit circle or cyclic groups , which do not obviously admit tensor product structure; this was first done by Bachelis, and with stronger quantitative estimates by Mockenhaupt-Schlag and by Green-Ruzsa.)

The tensor product trick is also widely used in additive combinatorics (I myself learned this trick from a survey paper of Ruzsa). Here, one deals with sets A rather than functions f, but the idea is still the same: replace A by the Cartesian power , see what estimate one gets, and let . There are many instances of this trick in the literature, but I’ll just describe one representative one, due to Ruzsa. An important inequality of Plünnecke asserts, among other things, that for finite non-empty sets A, B of an additive group G, and any positive integer k, the iterated sumset obeys the bound

(17)

(This inequality, incidentally, is itself proven using a version of the tensor power trick, in conjunction with Hall’s marriage theorem, but never mind that here.) This inequality can be amplified to the more general inequality

via the tensor power trick as follows. Applying (17) with , we obtain

The right-hand side looks a bit too big, but this is the same problem we encountered with the Cauchy-Schwarz or Holder inequalities, and we can resolve it in a similar way (i.e. by arbitraging homogeneity). If we replace G with the larger group and replace each set with the larger set , where is the standard basis for and are arbitrary positive integers (and replacing A with ), we obtain

Optimising this in (basically, by making the close to constant; this is a general rule in optimisation, namely that to optimise X+Y it makes sense to make X and Y comparable in magnitude) we obtain the amplified estimate

for some constant ; but then if one replaces with their Cartesian powers , takes roots, and then sends M to infinity, we can delete the constant and recover the inequality.

[*Update*, Sep 5: Optimal value for in the proof of Cauchy-Schwarz fixed. (Thanks to furia_kucha for the correction.)]

[*Update*, Sep 10: Proof of Ruzsa’s inequality fixed. (Thanks to Van Vu for the correction.)]

[*Update*, Sep 11: Depolarisation example added, and then corrected. (Thanks to Andy Cotton-Clay for the correction.)]

## 89 comments

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10 August, 2009 at 9:37 am

The Role of Amplifiers in Science « Gödel’s Lost Letter and P=NP[…] Tensor Power Trick This is a trick that is explained beautifully in Terry Tao’s great blog. See his post for the details, but here let me try to hopefully explain what is up. The […]

18 August, 2009 at 5:03 pm

The least quadratic nonresidue, and the square root barrier « What’s new[…] of two quadratic residues is again a quadratic residue. One way to use this sort of structure to amplify bad behaviour in a single short interval into bad behaviour across many short intervals. Because of […]

27 September, 2009 at 2:36 pm

The Gaussian Correlation Conjecture « Almost Sure[…] in the proof of 2 is a tensor power trick along the same lines as those described by Terence Tao in his blog. This will give the following result. Lemma 1 Let be positive constants with as goes to […]

18 November, 2009 at 6:31 am

Vorlesung Funktionalanalysis: Erste Etappe « UGroh's Weblog[…] die Cauchy-Schwarz Ungleichung. In diesem Zusammenhang möchte ich auf den Interessanten Artikel Amplification, Arbitrage and the Tensor Power Trick von T. Tao hinweisen, in dem die Methodik des Beweises der Cauchy-Schwarz Ungleichung in […]

2 December, 2009 at 2:48 pm

AJAre the techniques posted here of any relevance to entropy power inequalities arising in network information theory?

1 January, 2012 at 12:16 pm

Montgomery’s uncertainty principle « What’s new[…] which one dilates to (and replaces each frequency by their roots), and then sending (cf. the tensor product trick). But we will not need this refinement […]

29 March, 2012 at 9:50 am

Tomer ShalevHi Prof Tao,

regarding amplification of (17).

can you elaborate on the optimization of

it seems strange to pick a near global constant when one

knows that can be big.

29 March, 2012 at 10:36 am

Terence TaoSorry, by “constant” I meant “constant in i” rather than “absolute constant”. The point is that the arithmetic mean-geometric mean inequality is close to equality when the are comparable, no matter how large the are. In this particular application, we want to reverse the AM-GM inequality with , so the optimisation proceeds by setting the to be comparable, e.g. setting for some large parameter L.

29 December, 2012 at 1:04 pm

A mathematical formalisation of dimensional analysis « What’s new[…] to amplify a hybrid inequality into a dimensionally pure one by optimising over all rescalings; see this previous blog post for a discussion of this trick (which, among other things, amplifies the inhomogeneous Sobolev […]

17 January, 2013 at 6:15 am

JornAnother great example of amplification is the proof of Chernoff’s bound from Markov’s inequality in probability theory, which stunned me when I first came across it. Markov’s inequality states that $P(X \ge a) \le E(X)/a$ for non-negative random variables $X$. Using that $P(X \ge a) = P(\exp(\theta X) \ge \exp(\theta a))$ for any $\theta > 0$, we find that $P(X \ge a) \le E(\exp(\theta X)) \exp(-\theta a)$, in which the parameter $\theta > 0$ can be chosen freely.

4 March, 2013 at 6:45 am

Denis SerreHi Terry,

Great post! I like your style and try to follow it closely.

Here are two examples of amplification, where the amplification trick gets rid of constants.

1. The maximum principle for holomorphic functions in a disk (which could be another domain). Let and be a holomorphic function. Then Cauchy’s integral formula gives you an inequality , with respect to the sup-norm over the boundary. Apply the inequality to , then take the -th root. The constant is changed into , which tends to as tends to infinity.

2. In matrix analysis, let be an algebra norm over , that is . Let denote the spectral radius of a matrix (which involves the modulus of complex eigenvalues too). Then . This would be obvious if the norm was subordinated to a complex norm (take an eigenvector associated with an eigenvalue of maximal modulus, bla-bla). If not, take any such subordinated norm , we thus have . By equivalence of norms, we obtain . Apply this to , use , then take the -th root and let tend to infinity.

All the best,

Denis

10 May, 2013 at 10:05 pm

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10 May, 2013 at 10:06 pm

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23 July, 2013 at 5:33 pm

Cauchy-Schwarz Inequality (and Amplification) | Math ∩ Programming[…] the same. What makes this proof stand out is its insightful technique, which I first read about on Terry Tao’s blog. He calls it “textbook,” and maybe it is for an analyst, but it’s still very […]

4 November, 2013 at 1:49 pm

Real stable polynomials and the Kadison-Singer problem | What's new[…] case of the Kadison-Singer paving conjecture, but we may use a number of standard maneuvers to amplify this to the full paving conjecture, loosely following the papers of Popa and of Akemann-Weaver. […]

29 May, 2014 at 8:32 am

Inequalities Using Symmetries | Ah, Math?[…] by Terence Tao’s post about amplification of inequalities (https://terrytao.wordpress.com/2007/09/05/amplification-arbitrage-and-the-tensor-power-trick/) I will give a quick proof of the AM-GM inequality, which says that for any list of nonnegative […]

29 May, 2014 at 9:42 am

Inequalities Using Symmetries | Ah, Math?[…] by Terence Tao’s post about amplification of inequalities (https://terrytao.wordpress.com/2007/09/05/amplification-arbitrage-and-the-tensor-power-trick/) I will give a quick proof of the AM-GM inequality, which says that for any sequence of […]

15 February, 2015 at 5:03 pm

254A, Notes 3: The large sieve and the Bombieri-Vinogradov theorem | What's new[…] that the additional term could be deleted by an amplification trick similar to those discussed in this previous post). See this survey of Montgomery for these results and on the (somewhat complicated) evolution of […]

3 October, 2015 at 2:58 pm

275A, Notes 1: Integration and expectation | What's new[…] looks weaker than what we want to prove, but we can “amplify” this inequality to the full strength triangle inequality as follows. Replacing by for any […]

27 September, 2016 at 10:52 am

246A, Notes 2: complex integration | What's new[…] eliminate this loss we can amplify the above argument by exploiting phase rotation. For any real , we can repeat the above arguments […]

21 April, 2017 at 7:04 am

AnonymousCan this trick be used to show that (which is usually the second part of the textbook statement of the Cauchy inequality) implies that must be linearly dependent?

21 April, 2017 at 8:02 am

AnonymousA typical “textbook” proof (e.g. Folland’s Real Analysis, or https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Second_proof) of the Cauchy-Schwarz inequality goes like this:

In order to show that , one assumes without loss of generality . By expanding

$\displaystyle \|x-t\alpha y\|^2\geq 0$

where and , one can not only conclude that the Cauchy inequality is true

but alsothat the equality is true only when are linearly dependent.The choice of and seem rather mysterious; one the other hand, the way one chooses and in this excellent post is extraordinarily natural. However, maybe I miss something in the post; I don’t see how to conclude the “equality” part from your argument.

3 May, 2017 at 8:18 am

lowerboundAre any of these tricks available for lower bounds?

22 August, 2017 at 4:35 pm

An addendum to “arbitrage, amplification, and the tensor power trick” | What's new[…] one of the earliest posts on this blog, I talked about the ability to “arbitrage” a disparity of symmetry in an inequality, […]

24 August, 2017 at 11:20 am

AnonymousHi, Prof. Tao. I think there is a typo after the step of using ‘Khintchine’s inequality to compute the expectations’ on eq. (10). I think It should become

.

In the article, it missed raising the RHS to the power of .

Please correct me if I am wrong.

Thanks.

Anyway, thanks for this post. This is a really good post.

[Corrected, thanks – T.]4 October, 2018 at 12:30 pm

AnonymousDear Professor Tao

I am sorry for being of topic but this post opened my eyes on how to think about estimates. I am deeply grateful to you for the time and effort you put in writing these things. For people like me in 3rd world countries this is the only way to understand how to think about math and how to learn math.

Thank you very much

26 December, 2018 at 9:32 am

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12 January, 2019 at 7:37 am

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31 October, 2019 at 10:30 am

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27 January, 2020 at 10:13 am

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27 January, 2020 at 6:56 pm

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24 September, 2020 at 1:45 pm

Alan ChangIn the displayed equation following “We can then use the monotone convergence theorem to amplify the finite sum to an infinite sum, thus”, there is an extra power of p on the left-hand side.

[Corrected, thanks – T.]1 October, 2020 at 6:40 am

Bart SimpsonsLooks like WordPress doesn’t like the LaTeX here, it’s full of formula doesn’t parse unfortunately.

[Sometimes individual wordpress rendering servers have temporary glitches. Right now it seems to be rendering fine. -T]27 February, 2021 at 12:22 pm

Boosting the van der Corput inequality using the tensor power trick | What's new[…] While working on an ongoing research project, I recently found that there is a very simple way to get around the latter problem by exploiting the tensor power trick: […]