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Usually, at the heart of a good limit theorem in probability theory is at least one good inequality – because, in applications, a topological neighborhood is usually defined by inequalities. Of course, an explicit inequality may be even more useful by itself than its application to a limit theorem, which latter is in fact a statement about the mere existence of a certain kind of inequality (recall the presence of the quantifier $\exists$ in standard definitions of the limit).

In turn, a good inequality is oftentimes based on at least one good identity -- as something gets rewritten in some other, easier to analyze form. Identities used to prove inequalities are oftentimes simple and routine, such as $\int_a^c=\int_a^b+\int_b^c$. Among well-known examples of nontrivial and useful identities in analysis and probability are Cauchy's integral theorem, Plancherel's theorem, isometric imbeddings into $L^p$, duality identities of the $\max_x\min_y f(x,y)=\min_y\max_x f(x,y)$ type (under appropriate conditions on $f$), Spitzer's identity for random walks, and identities of Stein's method.

There are of course a great many other known identities (some of them nontrivial), such as thousands of formulas for integrals as e.g. in the Gradshteyn and Ryzhik collection. However, it appears that not many of those formulas can lead to interesting inequalities.

I think it would be useful for a number of people to learn about other, not widely known identities that can be/have been used to obtain interesting inequalities. A desirable answer to this question would state such an identity and indicate its possible/actual applications to inequalities, with appropriate references. Thank you!

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    $\begingroup$ I think this concern is probably too broad to be useful. Maybe I'll wait a while before voting in case people come up with really inspiring answers proving me wrong. At the very least, this should be community wiki. $\endgroup$ – Anthony Quas Aug 20 '15 at 21:24
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    $\begingroup$ I am actually afraid that the conditions of the question may be too narrow (restrictive); that is, there may be not too many not widely known identities that can be/have been used to obtain interesting inequalities. Of course, I would be glad if this apprehension is proven mistaken. $\endgroup$ – Iosif Pinelis Aug 20 '15 at 22:05
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I am not sure if this one fits to this category. In this case it is a PDE (or PDI -partial differential inequality) ruling the inequality. And (I think) you can extract some identity after ``integrating'' this PDE (or PDI).

Let $\Omega \subset \mathbb{R}^{2}$ be a rectangular subset, and let $H(x,y) : \Omega \to \mathbb{R}$ be a bounded smooth function such that $H_{x}, H_{y} \neq 0$ on $\Omega$. Fix some $\lambda \in (0,1)$. Then inequality $$ \int_{\mathbb{R}^{k}}\sup_{\lambda x+(1-\lambda)y=z}H(f(x),g(y)) d\gamma_{k}(z)\geq H\left(\int_{\mathbb{R}^{k}}f\; d\gamma_{k},\int_{\mathbb{R}^{k}}g\; d\gamma_{k} \right) $$ holds for all $(f(x),g(y)):\mathbb{R}^{k}\times \mathbb{R}^{k} \to \Omega$ where $d\gamma_{k}=e^{-x^{2}/2}\frac{1}{(2\pi)^{k/2}}dx$ is $k$ dimensional gaussian measure if and only if $$ (1-\lambda^{2}-(1-\lambda)^{2})\frac{H_{xy}}{H_{x}H_{y}}+\lambda^{2}\frac{H_{xx}}{H_{x}^{2}}+(1-\lambda)^{2}\frac{H_{yy}}{H_{y}^{2}}\geq 0 \quad (1). $$ Here is the reference to the formal statement

Applications:

You can try to play with some functions (or you can actually try to describe all solutions of PDE (1) -- when you have equality instead of inequality). For example, if you try $H(x,y)=\Psi(\lambda \Psi^{-1}(x)+(1-\lambda)\Psi^{-1}(y))$ then (1) takes the form $$ \frac{1}{\Psi'(\lambda P+(1-\lambda)Q)}\left(\frac{\Psi''(\lambda P+(1-\lambda)Q)}{\Psi'(\lambda P+(1-\lambda)Q)}-\lambda \frac{\Psi''(P)}{\Psi'(P)}-(1-\lambda)\frac{\Psi''(Q)}{\Psi'(Q)}\right), $$ where $P=\Psi^{-1}(x), Q=\Psi^{-1}(y)$, which is nonnegative if and only if $\Psi'>0$ and $\frac{\Psi''}{\Psi'}$ is concave (or $\Psi'<0$ and $\frac{\Psi''}{\Psi'}$ is convex).

1) (Prekopa--Leindler inequality) Indeed, take $\Psi(x)=e^{x}$ in the previous example. Then you will get $H(x,y)=x^{\lambda}y^{1-\lambda}$.

2) (Ehrhard inequality). Take $\Psi(t)=\int_{-\infty}^{t}\frac{e^{-x^{2}/2}dx}{\sqrt{2\pi}}$. This recovers Ehrhard inequality.

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  • $\begingroup$ It is a powerful tool, but what exactly is identity behind all this stuff? $\endgroup$ – Fedor Petrov Aug 22 '15 at 19:07
  • $\begingroup$ This seems a very interesting result. Even though the role of identities is not quite clear here, the case of the equality in (1) is already interesting. Is a proof of this result published? $\endgroup$ – Iosif Pinelis Aug 23 '15 at 16:04
  • $\begingroup$ @IosifPinelis, A rough version is uploaded on arxiv -- sorry for advertising :) $\endgroup$ – Paata Ivanishvili Aug 23 '15 at 19:24
  • $\begingroup$ @FedorPetrov, It is not quite clear for me. For example, I don't think there can be a simple identity for Prekopa--Leindler inequality as you have for Hardy inequality because characterization of extremizers is not ``unique''. But there are some identities involved behind this stuff in the paper in terms of matrices and semigroups. $\endgroup$ – Paata Ivanishvili Aug 23 '15 at 19:40
  • $\begingroup$ Is it possible to include the two instances of the function $H$ corresponding to the Prekopa--Leindler inequality and to the Ehrhard inequality into a continuous (say) family of functions $H$ for which the equality in (1) holds on $\Omega$? Then one would have a continuous family of inequalities containing the Prekopa--Leindler and Ehrhard inequalities. $\endgroup$ – Iosif Pinelis Sep 7 '15 at 15:52
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Many inequalities are proved by identities representing the thing which must be proved to be non-negative as integral (or sum, or expectation) of squares. For example: CBS inequality $$ \int_X f^2 \cdot \int_X g^2 -\left(\int_X fg\right)^2=\frac12\int_{X\times X} \left(f(x)g(y)-f(y)g(x)\right)^2\geqslant 0. $$ Or more involved: Hardy inequality $$4\int_0^\infty f^2(x)dx-\int_0^\infty \left(\frac1x \int_0^x f(t)dt\right)^2dx=\int_0^\infty\int_0^\infty (f(x)\sqrt{x}-f(y)\sqrt{y})^2\frac{dxdy}{2\sqrt{xy}\max(x,y)}$$

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  • $\begingroup$ This is not widely known? $\endgroup$ – Dirk Aug 22 '15 at 11:14
  • $\begingroup$ Well, my point is that it is less known than it deserves to be. I may be wrong. $\endgroup$ – Fedor Petrov Aug 22 '15 at 11:32
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Probably the Fenchel inequality counts? $\newcommand{\RR}{\mathbb{R}}$

For a function $f:X\to\RR\cup\{\infty\}$ on a vector space $X$ and it's convex conjugate $f^*(x^*) = \sup_{x\in X}\langle x^*,x\rangle - f(x)$ there always holds that $$\langle x^*,x\rangle\leq f(x) + f^*(x^*).$$

This include Cauchy-Schwarz and Young's inequality but can sometimes be used to get more specific inequalities.

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One example of identities that might fit the stated restrictions is the various Positivstellensätze of real algebraic geometry, which provide so-called certificates of positivity for polynomials $f(\mathbf x)\in\mathbb{R}[\mathbf x]$ of several real variables over semi-algebraic sets of the form $K:=\{\mathbf x\in\mathbb{R}^n\colon g_j(\mathbf x)\ge0\ \ \forall j\in J\}$, where $J$ is a finite set and $g_j(\mathbf x)\in\mathbb{R}[\mathbf x]$ for each $j$. Such a certificate is a representation of $f(\mathbf x)$ as a polynomial, with nonnegative coefficients, in the $g_j(\mathbf x)$'s and (sums of) the squares of polynomials in $\mathbb{R}[\mathbf x]$. See e.g. Lasserre. I used this method in BH ineq.

Of course, only polynomials $f(\mathbf x)$ that are actually nonnegative on $K$ can have certificates of positivity. What is remarkable is that, under mild conditions, all positive on $K$ polynomials $f(\mathbf x)$ have been shown to have such certificates -- which, however, may not be easy to obtain.

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In my paper, found at arxiv or AoP, the proof of an exact Rosenthal-type bound on absolute moments of sums of independent zero-mean random variables relied to a large extent on identities of "a calculus of variations", given by Lemma 2.5 there, of moments of infinitely divisible distributions with respect to variations of the corresponding Lévy characteristics (as well as on a number of other identities, such as the ones in display (71) there).

My apologies for giving another example related to my own work. I can try to justify this by saying that (i) my work is what I know the best, (ii) the question itself was rooted in my experience, and (iii) there have been not many answers so far.

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Another, very recent example of a proof of an inequality based on an apparently not-widely-known identity is my answer at An inequality concerning Lagrange's identity; cf. the answer by Fedor Petrov on this page concerning the Cauchy--Schwarz inequality.

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