17
votes

Accepted

### Find a special integer coefficients polynomial which takes small absolute value on [0,4]

No. Integer coefficients are a red herring; the point is that, if $f(x) = f_n x^n + \cdots + f_0$ is a real polynomial with $f_n \neq 0$, then the $L_{\infty}$ norm of $f(x)$ on an interval of the ...

8
votes

Accepted

### Inequality of inclusion-exclusion term

This is simply computable explicitly via Chu--Vandermonde convolution identity $$\sum_{a+b=n, a,b\geqslant 0}{x\choose a}{y\choose b}={x+y\choose n}.$$
Namely,
$$
(-1)^a \frac{(k-a-2)!}{a! (i-a-1)! (j-...

5
votes

### A question about the prime counting function

A numerical calculation (in, say, Wolfram Alpha or Mathematica) shows that if $x \geq e$, then
$$\frac{(\sqrt{x}+1)^2}{2\log(1+\sqrt{x})}<\frac{x}{\log x}\Big(1+\frac{1}{\log x}+\frac{2}{(\log x)^2}...

5
votes

Accepted

### Derive distributional inequalities from pointwise estimates

First note that the result is true if the support of $\varphi$ is disjoint from $E$; this follows from integration by parts.
Next note that every algebraic set is a finite union of smooth submanifolds ...

4
votes

Accepted

### Relating function value to $L^2$ norm in Holder space

Take any real $t\ge1$, so that $|f|\le1$ and $|f'|\le1$. Let $h:=|f(1/2)|$ and $y:=\|f\|_2$. Then $y\in[0,1]$, $h\in[0,1]$, and $|f(x)|\ge g(x):=\max(0,h-|x-1/2|)$ for all $x\in[0,1]$. So,
$$y^2\ge\...

4
votes

Accepted

### Is the passage in argument of existence solution of PDE correct?

This is probably a typo: probably the author meant $T<1$ here, instead of $T>1$. A bit later, in lines 3- and 2- at the bottom of the same page, it is explicitly assumed that $T$ is $<1$ and ...

4
votes

### Optimal constant comparing $f(1/2)$ and $\|f\|_2$ when $f$ is $t$-Hölder?

The following is a copy of my earlier answer which resolves this problem for integer $t$
Start with the Gagliardo-Nirenberg-Sobolev interpolation inequality, which in particular states that for ...

4
votes

### What is the minimum and the maximum perimeter of a triangle with area $x$ that can be inscribed in a circle?

The extrema are attained for symmetric (isosceles) triangles. Trigonometry gives for the pairs (area, perimeter) as a function of the half tip angle $\psi$ the relations
$$\left( 4 \,\sin \psi\,\cos^3\...

3
votes

### Does "perpendicular phase incoherence" satisfy the triangle inequality?

This follows from
Lemma. For complex numbers $x,y$ in the unit disk we have $$\sqrt{1-\Re xy}\leqslant \sqrt{1-\Re x}+\sqrt{1-\Re y}.\tag{1}$$
Proof. Put $x=u+iv$,$y=p+iq$. Then, with fixed $p,u$ and ...

3
votes

### Is the passage in argument of existence solution of PDE correct?

It is a typo.
Look at the bottom of the same page: where the author says to choose "$T = T(p,n,l,\|\phi\|_X) < 1$ such that ..."
The author is proving local existence here, and it makes ...

3
votes

Accepted

### Erdős–Rényi random graphs — reproducing 2 inequalities

$\newcommand{\lln}{\operatorname{\lln}}\newcommand{\bb}{\binom}$We are going to deduce from \eqref{1} a bound better than the ultimate bound in \eqref{2}. Actually, we are going to obtain the exact ...

3
votes

Accepted

### Explicit upper and lower bounds for a certain support function

Let $$g_t(a) = \max_{1\le k\le n}\ \min\left(\sqrt{\sum_{i=1}^k a_i^2},\ t^2\frac{\sum_{i=1}^k a_i^2}{\left(\sum_{i=1}^k\sqrt{a_i}\right)^2}\right).$$ Then $g_t(a) \le f_t(a) \le 3g_t(a)$. The proof ...

3
votes

Accepted

### Ratio of the constants of the Marcinkiewicz–Zygmund inequality for p=1

You are missing the crucial independence and zero-mean conditions, without which the Marcinkiewicz–Zygmund inequalities will not hold in general.
These inequalities are actually as follows: for each ...

2
votes

### Recover unknown vector through shifted argmax queries

The following argument suggests that the factor $O(n)$ cannot be improved. Suppose for a contradiction that $x\in\{0,1\}^n$ and after queries with $v_1,\dots,v_{n-1}\in\mathbb R^n$ we determined that $...

2
votes

### Most elementary proof showing that exponential growth wins against polynomial growth

For $k\ge1 $ and $$n\ge 4k^2$$ (that is $\frac{2k}{\sqrt n}\le 1$) set $x_1=\dots=x_{2k}=\sqrt n$ and $x_{2k+1}=\dots=x_n=1$. The arithmetic mean of these $n$ numbers is $\frac{ 2k\sqrt n +n-2k}n\le \...

Community wiki

1
vote

### Relating function value to $L^2$ norm in Holder space

Edit
For $t > k \in \mathbb{N}$, we can get the conjectured rate. I leave my original non-sharp answer below the cut.
Start with the Gagliardo-Nirenberg-Sobolev interpolation inequality, which in ...

1
vote

Accepted

### Maximizing a sum minus its maximal summand

It is true. The proof rests on several observations.
The first one is that if you want to maximize $\sum_i i\pi_i$ under the condition $\pi_j\le B_i$ with any prescribed $B_i$, then your best bet is ...

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