Identities and inequalities in analysis and probability

Question

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!

Answer 1

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.

Answer 2

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 $$4\int_0^\infty f^2(x)dx-\int_0^\infty \left(\frac1x \int_0^x f(t)dt\right)^2dx=\int_0^\infty\int_0^\infty (f(x)\sqrt{x}-f(y)\sqrt{y})^2\frac{dxdy}{2\sqrt{xy}\max(x,y)}. \tag{$\heartsuit$}$$ To prove $(\heartsuit)$, we may "expand the brackets" and look at "coefficients" of $f^2(t)$ and $f(t)f(s)$: in LHS we have the expression $$ \int_0^\infty \left(\frac1x \int_0^x f(t)dt\right)^2dx= \int_0^\infty \frac{dx}{x^2} \int_0^x f(t)dt \int_0^x f(s)dsdx\\ =\int_0^\infty dt\int_0^\infty ds\, f(t)f(s)\int_{\max(t,s)}^\infty \frac{dx}{x^2}=\int_0^\infty dt\int_0^\infty ds\, \frac{f(t)f(s)}{\max(t,s)}, $$ which is cancelled with the part of RHS containing $f(x)f(y)$. For the part of RHS containing $f^2(x)$ or $f^2(y)$ (these two integrals are equal to each other by symmetry), they two give $$ \int_0^\infty dx f^2(x) \int_0^\infty \frac{\sqrt{x}}{\sqrt{y}\max(x,y)}dy, $$ the inner integral equals 4, as in LHS.

Answer 3

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.

Answer 4

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.

Answer 5

I found the following elementary identity quite useful in proving some non-trivial inequalities: $$ \min\{u,v\} = \sqrt{uv}\exp\left(-\frac12\left|\log\frac{u}{v}\right|\right), \qquad u,v>0. $$

Based on this identity, I was able to derive the following fact. Let $P_i,Q_i$, $i\in[n]$, be distributions on a finite set $\Omega$. We will use $P^\otimes_{i\in[n]}$ to denote $n$-fold products of distributions and $||\cdot||$ will denote the total variation norm (i.e., half the $\ell_1$ norm). Let $$ \delta=(\|P_1-Q_1\|,\|P_2-Q_2\|,\ldots,\|P_n-Q_n\|) $$ be the vector of marginal total variation distances.

We have the obvious relation

$$ \|\delta\|_\infty \le \| P^\otimes_{i\in[n]}-Q^\otimes_{i\in[n]} \| \le \min\{1,\|\delta\|_1\}. $$

In a recent preprint https://arxiv.org/abs/2409.10368, I sharpened the lower bound to $$ \| P^\otimes_{i\in[n]}-Q^\otimes_{i\in[n]} \| \ge c\min\{1,\|\delta\|_2\} $$ for some universal constant $c>0$.

Answer 6

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.

Answer 7

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.