$L^P$ spaces inequality

Question

I would like to prove:

Suppose $X\subset\mathbb{R}^m$ and $Y\subset\mathbb{R}^n$ are measurable sets, and $1\leq p\leq \infty.$ Show that if $f(x,y):X\times Y \to \mathbb{R}$ is measurable and nonnegative, then $$\left\|\int_Y f(x,y)dy\right\|_{L^P(X)}\leq \int_Y\|f(x,y)\|_{L^P(X)}dy$$

I would greatly appreciate a hint to prove this result. I tried to use the theorem that if $$\sup\left\{\int_E fg:g\in L^q(E),\|g\|_L^q(E)\leq1\right\} \leq M <\infty,$$ then $f\in L^P(E)$ and $\|f\|_{L^P(E)}\leq M.$ But I can't see how to proceed from here.

2

You can find the proof here – Nyfiken Nov 13 '16 at 21:24

Alessandro Blasetti · Answer 1 · 2016-11-13T21:48:32.140

1

Let $p=\infty$, using Jensen inequality we get: $$ \left\|\int_Y f(x,y)dy\right\|_{L{(X)}^\infty}=\sup_X\left|\int_Yf(x,y)dy \right|\le\sup_X\int_Y\left|f(x,y)\right|dy\le\int_Y\sup_X\left|f(x,y)\right|dy=\int_Y\left\|f \right\|_{L{(X)}^\infty}. $$ The case $1\le p< \infty $ is given in a comment to the question.

edited Nov 13 '16 at 21:48

answered Nov 13 '16 at 21:05

Alessandro Blasetti

598

It looks like you have missed some power $p$ (or $1/p$ resp.) in your estimations. – Nyfiken Nov 13 '16 at 21:23
@Nyfiken , you are right. I am thinking how to correct the inequalities. Feel free to correct if you already know how! – Alessandro Blasetti Nov 13 '16 at 21:40
I do not know a short proof. I gave two links above in the comment under OP that have some suggestions. – Nyfiken Nov 13 '16 at 21:45
Edited my answer – Alessandro Blasetti Nov 13 '16 at 21:46

$L^P$ spaces inequality

1 Answers1