# Norm of the operator induced by the essential bounded function

Consider a measure space $$(\Omega,\mathcal{F},\mu)$$. Let $$\alpha\in L^\infty$$. Here $$L^\infty$$ means $$L^\infty(\Omega,\mathcal{F},\mu)$$. Let $$T\colon L^p\to L^p, f\mapsto \alpha\cdot f$$ for some fixed $$p\in(1,\infty)$$. Find the norm $$\lVert T\rVert$$.
My thoughts: I want to prove $$\lVert T\rVert =\lVert \alpha\rVert_\infty$$ since this is a natural norm and I have proved $$\lVert T\rVert \leq\lVert \alpha\rVert_\infty$$. To get the reversed inequality, I supposed that $$\mu$$ is a finite measure and the inequality follows from Chebyshev's inequality (considering the set $$E_\varepsilon=\{\omega\in\Omega: |\alpha(\omega)|>\lVert\alpha\rVert_\infty-\varepsilon\}$$ and $$\chi_{E_\varepsilon}\in L^{p}$$).
If $$\mu$$ is a $$\sigma$$-finite measure, I think the inequality can be proved similarly: let $$A_1\subseteq A_2\subseteq \cdots \subseteq A_n\subseteq \cdots$$ be a sequence of sets with finite measure and consider the sets $$A_j\cap E_\varepsilon$$. Then the monotone convergence theorem works.
I wonder: In which ways can we prove the reversed inequality without assumption on $$\mu$$, or is there some counterexamples?

• This post has discussed the $\sigma$-finite case. Commented Oct 3, 2022 at 6:57
• If $\mu (A)=\infty$ for every non-empty set $A$ the $L^{p}=\{0\}$ and $T=0$ but $\alpha\|_{\infty}$ is arbitrary. Commented Oct 3, 2022 at 7:13
• @geetha290krm Yes, I proved this using the Chebyshev's inequality. But how does this result work here? Commented Oct 3, 2022 at 7:32
• @geetha290krm thank you, I think I have done. Commented Oct 3, 2022 at 10:45