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Let $(\Omega, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $g$ a measureable function on $\Omega$. Fix $p\in[1,\infty]$ and consider the multiplication operator $$M_g:L^p(\Omega, \mu)\to L^p(\Omega, \mu),\quad f\mapsto fg$$ Then $M\in B(L^p(\Omega, \mu))$ if and only if $g \in L^\infty(\Omega, \mu)$ and in that case $\|M\| = \|g\|_{L^\infty}$.

My attempt:

I proved the"$\impliedby$" direction.

"$\implies$"Suppose $M$ is a bounded operator and assume that $g$ is not essentially bounded. Then there exists a set $A$ of nonzero measure such that $g$ is unbounded on $A$. By $\sigma$-finiteness, we can choose an increasing sequence $(A_n)\subset \mathcal A$ with $\bigcup_n A_n=\Omega$. I tried considering the functions $f_n=\mathbf 1_{A\cap A_n}$ to get a contradiction, but my attempts have failed.

I'm also not sure how to prove $\|M\| = \|g\|_{L^\infty}$. I'm fine with the inequality $\|M\|\leq \|g\|_\infty$, but I don't know how to show the other one.

Thank you for any help.

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    $\begingroup$ If $g $ is not essentially bounded, then $B_n =\{x \mid |g (x)|>n\} $ has positive measure for each $n $. Now consider the functions $f_{n,m}=1_{A_n \cap B_m} $, noting that $A_n \cap B_m $ has positive measure for a suitable $n=n (m) $ (why). $\endgroup$
    – PhoemueX
    Commented Oct 31, 2015 at 15:16

1 Answer 1

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  1. If $g\in L^{\infty}$, the it is clear that $M \in B(L^p)$ and $\|M\| \leq \|g\|_{\infty}$. Now consider the sets $$ E_n := \{x : |g(x)| > \|g\|_{\infty} - 1/n\} $$ Then $E_n$ has positive measure, so $\exists m\in \mathbb{N}$ such that $E:= E_n\cap A_m$ has positive measure. Let $f := \chi_E$, then $f\in L^p$ and $$ \|f\|_p = \mu(E)^{1/p} $$ Also $$ \|M\|\mu(E)^{1/p} \geq\|Mf\| = [\int|gf|^p]^{1/p} > (\|g\|_{\infty} - 1/n) \mu(E)^{1/p} $$ And hence $\|M\| > \|g\|_{\infty} - 1/n$ for all $n$, proving that $$ \|M\| = \|g\|_{\infty} $$

  2. Now suppose $M \in B(L^p)$, then use @PhoemeuX's hint to prove that $\exists n\in \mathbb{N}$ such that $|g| \leq n$ almost everywhere.

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  • $\begingroup$ Thank you very much, I see it now! $\endgroup$ Commented Nov 2, 2015 at 9:31

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