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for questions involving inequalities, upper and lower bounds.

4 votes
1 answer
232 views

Optimal constant in Carleman's inequality

Let $(a_j)_{j\ge 1}$ be a sequence of positive real numbers. Carleman's inequality says that $$ \sum_{n\ge 1}\left(\prod_{1\le j\le n} a_j\right)^{1/n}< e\sum_{n\ge 1} a_n. $$ The constant $e$ is opti …
Bazin's user avatar
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2 votes
1 answer
200 views

Sobolev spaces based on $L^p$ with $0<p<1$

I am looking for references on the topic of Sobolev spaces based on $L^p$ with $0<p<1$. For instance, a natural question could be: let $u$ be a (compactly supported) distribution on $\mathbb R^n$ suc …
Bazin's user avatar
  • 16.2k
2 votes

Is the 2D Ladyzhenskaya inequality true for periodic functions?

Let us start with the so-called Gagliardo-Nirenberg Inequality in $n$ dimensions, $$ \Vert u\Vert_{L^{n/(n-1)}(\mathbb R^n)}\le c_n\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag{GN}$$ an inequality th …
Bazin's user avatar
  • 16.2k
1 vote

Local fractional Sobolev inequality

For $\epsilon >0$, $u\in H^{\frac{n}{2}+2\epsilon}$, $N_0=-\Delta+1$, $$ \Vert N_0^{\frac{n}{4}+\epsilon} u\Vert_{L^2}=\Vert u\Vert_{H^{\frac{n}{2}+2\epsilon}}\ge c_{n,\epsilon} \Vert u\Vert_{L^\inf …
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  • 16.2k
2 votes
Accepted

An inequality involving multi-index

The first inequality with constant 1 follows from $ \vert x^\alpha\vert\le\Vert x\Vert_{\infty}^{\vert \alpha \vert},\quad\text{where $\Vert x\Vert_{\infty}$ is the sup-norm.} $ The second equality, …
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  • 16.2k
0 votes

Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\...

Too long for a comment. Let us consider for $X\in \mathbb S^{n-1}$, $ \langle X,(e^{i \alpha k})_{1\le k\le n}\rangle_{\mathbb C^n}. $ The question at hand is $$ \max_{X\in \mathbb S^{n-1}}\vert\langl …
Bazin's user avatar
  • 16.2k
1 vote

A generalization of discrete Hilbert's transform (Montgomery's inequality)

Let me deal with a continuous situation. Let $\lambda:\mathbb R\rightarrow\mathbb R$ be an increasing $C^1$ diffeomorphism and let $u,v$ be in $L^2(\mathbb R)$. We have with $\phi=\lambda^{-1}$, $$ A= …
Bazin's user avatar
  • 16.2k
1 vote
2 answers
221 views

Logarithmic Sobolev embeddings

Is there a clear connection with Gross' logarithmic Sobolev inequalities? …
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  • 16.2k
0 votes

Logarithmic Sobolev embeddings

Following Christian Remling suggestion, it seems that $$ \Vert u\Vert_{L^\infty(\mathbb R^d)}\le \gamma(d)\bigl\{ \Vert u\Vert_{L^2(\mathbb R^d)}+ \Vert \vert D\vert^{d/2} L(\vert D\vert) u\Vert_{L^2( …
Bazin's user avatar
  • 16.2k
5 votes
1 answer
371 views

Generalization of Gagliardo-Nirenberg Inequality

The standard Gagliardo-Nirenberg Inequality is $$ \Vert u\Vert_{L^{\frac{n}{n-1}}(\mathbb R^n)}\le C_n \Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag{$\ast$}$$ and constitutes a key step to proving So …
Bazin's user avatar
  • 16.2k
8 votes
1 answer
530 views

Inequality for Laguerre polynomials

Let $L_n$ be the $n$-th Laguerre polynomial defined by $\quad L_n (x)=\frac{e^x}{n!}\frac{d^n}{dx^n}(x^n e^{-x}).\quad $ I want to prove that $$ \forall n\in \mathbb N,\forall x\ge 0,\quad \sum_{0\le …
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  • 16.2k
2 votes

Is $1/\max(i,j)$ a bounded matrix on Hilbert spaces?

On $\ell^p(\mathbb N^*)$, $1<p<+\infty$, the matrix $$ A=\left(\frac{1}{i+j}\right)_{1\le i,j},\tag 1 $$ is a bounded operator and since the entries are positive and "equivalent" to your matrix, the …
Bazin's user avatar
  • 16.2k
0 votes
Accepted

An interpolation inequality.

$$ g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k-1}2\ln(1-\epsilon)} \lesssim \int_0^{+\infty} x^s e^{-a\epsilon x} dx= \int_0^{+\infty}x^s e^{-ax}dx\epsilon^{-s-1} $$ where $a$ is a fixed constant. So $$ …
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  • 16.2k
2 votes

Gronwall estimate with a Fourier transform

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=\al …
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  • 16.2k
4 votes

(sharp)Garding's inequality and inequality with lower bounds

Let us start with a short review of the so-called G{\aa}rding's inequality. Let $a\in S^m$ be nonnegative. Then with $a^w$ standing for the Weyl quantization of the Hamiltonian $a$, $$ (1)\quad m=1\qu …

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