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Prove that: if $g \in L^{\infty}$, the operator $T$ defined by $Tf = fg$ is bounded on $L^{p}$ for $1\leq p\leq \infty$. Its operator norm is at most $||g||_{\infty}$, with equality if $\mu$ is semifinite, where $\mu$ is the measure on $\mathcal{M}$, the measure space.

My approach: I consider $\hat{g(x)}(f) = f(g(x))$, which is a linear operator on $L^{p}$. Clearly $||\hat{g(x)}|| = ||g(x) || \leq ||g||_{\infty}$ , which gives me the first part of the proof. I am clueless about the second part, involving semifinite measure.

  • 0
    How are you computing $f(g(x))$? $g$ takes values in $\mathbb R$ (or $\mathbb C$, I guess) while $f$ is defined on $\mathcal M$...2012-03-07
  • 0
    $g$ is in $L^{\infty}$2012-03-07
  • 4
    $\mathcal M$ might be the set of chairs in Asia, for all you know...2012-03-07

3 Answers 3