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Let X be sigma finite measure space and $\phi$ $\in$ $L^{\infty}(X, \Omega)$ and $M_\phi:L^p(X, \Omega)$ $\to$ $L^p(X, \Omega)$ multiplication operator then show that $\| M_\phi \|=\| \phi\|_{\infty}$.

My attempt:

I could prove that $\| M_\phi \| \leq \| \phi \|_ {\infty}$ And then to prove the reverse inequality, I tried constructing a sequence of functions $f_n$ in $L^p$ such that $\| \phi f_n \|_p$ converges to $\| \phi \|_ \infty$ in the field. If we get such a sequence then that will prove the result by property of sup. Can we construct such a sequence?

As X is sigma finite so $X=\cup_{i=1}^{\infty} X_i$ where $\mu(X_i) < \infty$

I tried defining $f_n(x)=1 $ if $x \in \cup_{i=1}^n X_i$ and $0$ otherwise. But then the norm of $M_\phi(f_n)$ converges to $\int \phi^p$

Can i modify this? Or is there any other way?

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You might first try the (special) case with vectors, i.e., equip $\mathbb{R}^n$ with the $\ell^p$-norm, i.e., $\|x\|_p^p = \sum_{i=1}^n |x_i|^p$. How do you have to choose $x$ to obtain $$\| x \, \phi \|_p = \|x\|_p \, \|\phi\|_\infty\qquad?$$

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  • $\begingroup$ i have posted ans for my qn. @gerw can you pls check it? $\endgroup$ – Meow Feb 8 '17 at 16:25
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I think i have solved it. Can someone please check it..

$\| \phi \|_{\infty} = inf$ $\lbrace c>0 : \mu(\lbrace x\in X:|\phi(x)|>c\rbrace)=0\rbrace$

let $C_X=\big\lbrace c>0 : \mu(\lbrace x\in X:|\phi(x)|>c\rbrace)=0\big\rbrace$

now clearly $\| \phi \|_{\infty}- \epsilon \notin C_X$ So $\mu(\lbrace x\in X:|\phi(x)|>\| \phi \|_{\infty}- \epsilon \rbrace)\neq 0$ And using sigma finiteness of X from the above set we can get a set D whose measure is finite non zero and on which $|\phi(x)|>\| \phi \|_{\infty}- \epsilon $. Then define $f= \dfrac{\chi_D}{\mu(D)^{1/p}}$ Then for this $f$ its easy to prove that it is infact in $L^P$ and $\|M_\phi(f)\| \geq \| \phi\|_\infty-\varepsilon$ and letting $\varepsilon$ tends to zero we get the required result.

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  • $\begingroup$ One point (easy to fix, but important): $\int_D \phi\,d\mu$ can be $0$, you need to modify your $f$ so that $f\cdot \phi$ has constant sign [or constant argument, if we're looking at complex valued functions] (a.e.). But the idea is correct. $\endgroup$ – Daniel Fischer Feb 8 '17 at 16:36
  • $\begingroup$ didnt get what u r saying. ny ways $f \phi$ will be inside modulus $\endgroup$ – Meow Feb 9 '17 at 4:34
  • $\begingroup$ D'oh. Sorry, I thought of $f \mapsto \int f\phi$ when I wrote that. For the multiplication operator on $L^p$ you're right of course, no modification necessary. $\endgroup$ – Daniel Fischer Feb 9 '17 at 12:19

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