24

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 ...

answered Jun 20 at 15:10

Noam D. Elkies

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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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