Interesting inequality $\|F\|_p\le \frac{\pi}{\sin(\pi/p)}\|f\|_p$ over $L^p$

Question

Consider the function $$F(x)=\int_0^\infty \frac{f(y)}{x+y} \, dy, \quad0<x<\infty$$

Prove that if $1<p<\infty$, $$\|F\|_p\le \frac{\pi}{\sin(\pi/p)}\|f\|_p$$ and show that the constant is the best possible.

Since this problem came from a chapter on convolution, I think it might help to rewrite the integral in the form of a convolution over $\mathbb{R}$. I'm thinking of setting $z=x+y$ so that $y=z-x$.

Also, to evaluate the constant, this might be helpful:

for $0<a<1$, $$\int_0^\infty \frac1{(1+x)x^a} \, dx=\frac{\pi}{\sin\pi a}$$

Somewhat related: Prove $ F(x)=\int_0^{\infty}\frac{f(y)}{x+y}dy $ is continuous on $(0,\infty)$ and differentiable, and have $\lim\limits_{x\to \infty} F(x)=0$.

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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From this answer we know that $$ \Vert F\Vert\leq \int_0^\infty\frac{1}{(1+t)t^{1/p}}dt $$ From this answer we know that the last integral equals $$ \frac{\pi}{\sin(\pi/p)} $$

edited Apr 13 '17 at 12:20

Community

1

answered Nov 10 '13 at 07:07

Norbert

56,803

Thank you! But how do we show the constant is the best possible? – Christmas Bunny Nov 10 '13 at 19:38
This wa not asked in the original question. You can try this approach but it heavily uses that $p=2$. – Norbert Nov 10 '13 at 20:08

Interesting inequality $\|F\|_p\le \frac{\pi}{\sin(\pi/p)}\|f\|_p$ over $L^p$

1 Answers1