# Thread: Help with bounded multiplication operator proof

1. ## Help with bounded multiplication operator proof

Hi guys, i need to prove that the operator Tf = x*f(x) maps from L^p(-2,2) into L^p(-2,2) where:
$L^p(-2,2) = \lbrace f: [-2,2] \rightarrow \mathbb{C}:\, \int\limits_{-2}^2 |f(x)|^p\,\text{d}x < \infty \rangle$
I think that Holders inequality would be useful so i use it like this
$\int\limits_{-2}^2 |f(x)x|^{p-1}|f(x)x|\,\text{d}x \leq \int\limits_{-2}^2 \left(|f(x)x|^{p-1}\left[\int\limits_{-2}^2|f(x)|^p\, dx\right]^{1/p}\left[\int\limits_{-2}^2|x|^q\right]^{1/q}\right)dx = ||f(x)||_p \int\limits_{-2}^2 \left(|f(x)x|^{p-1} \int\limits_{-2}^2 |x|^q\right]^{1/q}\right) dx
$

By doing that until the integrand of the order p equals 1 i obtain with the relation $1/p+1/q = 1$:
$16 ||f(x)||_p^p 2^{p/(p-1)}(2p-1)^{1-p}(p-1)^{p-1}$
This expression converges when p goes to infinity so can i conclude that it maps as wished?
But i cant assume that the norm^P is finite can i?
Is there a smarter way(correct)
Im really lost, fought this problem for almost 6 hours.

2. ## Re: Help with bounded multiplication operator proof

Since you know Hölder's inequality the argument is easy, and in fact a more general result can be obtained in the same way: Let $g\in L^\infty(-2,2)$ then the mapping $f\mapsto gf$ is a bounded linear operator $L^p(-2,2)\to L^p(-2,2)$ for any $1\leq p\leq \infty$. To see this (the case $p=\infty$ is trivial, so assume it's finite) just note that Hölders inequality (with exponents $1,\infty$ associated to $|f|^p, |g|$ respectively) give $\| gf\|_p \leq \| g\|_{\infty}\| f\|_p$