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The class of concentration of measure inequalities is a fundamental tool in modern probability (and any field that uses probability, e.g., random matrix theory, theoretical computer science, statistics, high-dimensional geometry, combinatorics, etc.). As explained in this blog post of Scott Aaronson, these are basic ways in which one "upper bounds the ...

15

Gaussian Jensen's inequality: Let $\boldsymbol{X}=(X_{1}, \ldots, X_{n})\sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol\Sigma)$ be a gaussian vector. The inequality $$\mathbb{E} B(f_{1}(X_{1}), \ldots, f_{n}(X_{n})) \leq B(\mathbb{E}f_{1}(X_{1}), \ldots, \mathbb{E}f_{n}(X_{n}))$$ holds for all real valued (test functions) $f_{1}, \ldots, f_{n}$ if and only ...

9

An LMFDB search for curves with many integer points turns up the curve 20888a1: $y^2 = x^3 - 52 x + 100$ which has $52$ integral points, a bit more than your conjectured bound of about $47.052$ using $(M_x, M_y) = (12214, 1349854)$. It does seem to be true that curves with many integer points tend to have a few large ones, but I don't know of any ...

8

$$\|\sqrt{A+B}Cx\|^2=(\sqrt{A+B}Cx,\sqrt{A+B}Cx)=\\ ((A+B)Cx,Cx)=(ACx,Cx)+(BCx,Cx)=\|\sqrt{A}Cx\|^2+\|\sqrt{B}Cx\|^2,$$ taking the supremum over unit vectors $x$ we get $$\|\sqrt{A+B}C\|^2\leqslant \|\sqrt{A}C\|^2+\|\sqrt{B}C\|^2\leqslant (\|\sqrt{A}C\|+\|\sqrt{B}C\|)^2.$$

7

Write $$G=\left( \begin{array}{ccc} a & b & c \\ b & d & e \\ c & e & f \\ \end{array} \right)$$ and, without loss of generality, $$U=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & u & 0 \\ 0 & 0 & v \\ \end{array} \right),$$ where $u,v>0$. Then $$\text{Tr}(GUGU^{-1})-\text{Tr}(G^2) =\frac{b^2 (u-1)^2 v+... 6 Strichartz estimates, which originated from Strichartz, Robert S., Restrictions of Fourier transforms to quadratic surfaces and decay of solutions of wave equations, Duke Math. J. 44, 705-714 (1977). ZBL0372.35001, are a family of inequalities that provide L^p (or Sobolev) type control of solutions of linear dispersive or wave equations (such as the ... 6 This is from Garding. Let P\in{\mathbb R}[X_1,\ldots,X_d] be a homogeneous polynomial. Assume that it is hyperbolic in some direction e\in{\mathbb R}^d (with say the normalisation P(e)=1) and let \Gamma be its cone of future, that is the connected component of e in the complement of \{P=0\}. It is known that \Gamma is convex. Then we have the ... 6 For any unit vector x\in L^2(\mathbb{R}^n),$$\|\sqrt{A+B}\,Cx\|^2=(\sqrt{A+B}\,Cx,\sqrt{A+B}\,Cx) =(ACx,Cx)+(BCx,Cx)=\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|^2 \le(\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|)^2 \le(\|\sqrt A\,C\|+\|\sqrt B\,C\|)^2.$$So,$$\|\sqrt{A+B}\,C\|\le\|\sqrt A\,C\|+\|\sqrt B\,C\|.$$4 For any real numbers u,v,c such that u\le c\le v, let \mu_{c;u,v} denote the unique probability distribution on the set \{u,v\} with mean c. Your generalization of Jensen's inequality follows immediately from the well-known fact that any probability distribution \mu on \mathbb R with a given mean c\in\mathbb R is a mixture of probability ... 4 I came across this inequality on the Gaussian space recently. I was not aware of its existence since it is not really a classical one in comparison to the Poincaré inequality or the Logarithmic Sobolev inequality but it seems to be useful in order to prove the analyticity of the Ornstein-Uhlenbeck semigroup in L^p(\gamma). Let \gamma be the standard ... 4 Suppose we have such an n where P(n) is false. Now define (a, b, c) := (1, n(n-2), (n-1)^2) and observe that these three numbers are pairwise coprime and satisfy a + b = c. Then we have:$$ \textrm{rad}(abc) = \textrm{rad}(n(n-1)^2(n-2)) = \textrm{rad}(n(n-1)(n-2)) \leq n $$Moreover, the radical of n(n-1)(n-2) cannot be equal to any of n, n-1... 3 A really simple but powerful inequality is the so-called improved Kato inequality. I first learned about it when I was studying Uhlenbeck's removable singularity theorem for self-dual Yang-Mills connections. However, when I explained the inequality to Duong Phong and Eli Stein in Phong's office, Stein reacted with "It's in my book! It's in my book!"... 3 Concerning your first question: Let p_k:=P(B_{n,p}=k). We have to show that \begin{equation*} p_k-p_{k+1}\ll\frac1{npq}, \tag{1} \end{equation*} where q:=1-p and a\ll b means that a\le Cb for some universal real constant C>0. Clearly, without loss of generality (wlog) \begin{equation*} 1\ll npq. \end{equation*} Since p_{k+1}=\frac{n-k}{k+... 3 Esseen's anti concentration inequality is the basis of a lot of relatively recent (past 10-15 years) work on non asymptotic random matrix theory, particularly results on the smallest singular values of many random matrix models. It states that if Y is a real valued random variable, then$$\sup_{t \in \mathbb{R}} \mathbb{P}(|Y-t| \le 1)\le \int_{-2}^2 |\...

2

The expander Chernoff bound is a particularly nice generalization of the Chernoff inequality that is not so well known. It states the following: Let $G = (V,E)$ be a regular graph and consider a function $f : V \rightarrow [0,1]$. Perform a random walk $v_1, \cdots, v_t$ on $G$ by first picking $v_1$ uniformly at random. Then $$\mathbb{P}\left(\frac{1}t \... 2 I will assume c\le 2, since that seems to be the case you are interested in. The argument below is formulated in terms of taking a given c and computing the optimal \alpha, but this is equivalent to your question and the result agrees with your conjecture. Step 1: We can change the formulation of the question so that the weights can be any nonnegative ... 2 I will more comfortable with the notation v_\epsilon=\hat{u_\epsilon}; you have then$$ v_\epsilon(t,x)=\alpha(t,x)+\int_0^t\int e^{2πix(\xi+\epsilon\phi(s,\xi))} \hat{v_\epsilon}(s,\xi) d\xi ds=\alpha(t,x)+\int_0^t \bigl(\textrm{Op}(e^{2πix\epsilon\phi(s,\xi)}) v(s,\cdot)\bigr) (x) ds $$where \textrm{Op}(e^{2πix\epsilon\phi(s,\xi)})=A_{\epsilon, s} is ... 1 We have 0\le a\le b\le1 and T\in(0,\infty). We want to know when there is a positive constant C such that$$\int_0^T dt\, \int_a^b dx\, u^2(x-t)\geq C\int_0^1 dx\,u^2(x) \tag{1}$$for all measurable functions u\colon\mathbb R\to\mathbb R such that u(x)=0 for x\notin(a,b). The answer is: never. Indeed, without loss of generality a<b. The left-... 1 Let l(a,b) denote the left-hand side of both inequalities. Then l(b,a)=l(a,b)=l(-a,-b). The right-hand sides of both inequalities have the same properties. So, without loss of generality (wlog) a\ge b and a\ge0. Also, by homogeneity, wlog a=1. So, we have these two cases to consider:$$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1. ...

1

I am not a professional mathematician, thus feel free (you or your colleagues/professors of the site) tell me if my answer doesn't fit with your requirements that I can to delete it. An important inequality in complex analysis and functional analysis is the statement of Hadamard three-lines theorem, see the section Statement from this link of the Wikipedia ...

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