Boundedness of the multiplication operator

Let $$(\Omega,\mathcal{F},\mu)$$ be a measure space and $$f: \Omega \to \mathbb{C}$$ be a measurable function. Let $$M_f$$ be the multiplication operator whose domain is those $$g\in L^2(\Omega,\mathcal{F},\mu)$$ for which $$fg \in L^2(\Omega,\mathcal{F},\mu).$$

If $$f$$ is essentially bounded, then I can prove that $$M_f$$ is a bounded operator with $$\|M_f\|=\|f\|_{\infty}$$ where $$\|f\|_{\infty}$$ is the essential supremum of $$f.$$

I am unable to prove the converse! Here is my attempt.

Suppose $$f$$ is not essentially bounded, then $$E_n=\{w \in\Omega: |f(w)| \geq n\}$$ has positive measure for all $$n \in \mathbb{N}.$$ So, if I can find for every $$n$$ a non-zero $$g$$ in the domain of $$M_f$$ that vanishes outside $$E_n$$ then we get \begin{align*} \|M_f(g)\|^2&= \int_{\Omega} |fg|^2 \\ &= \int_{E_n} |f|^2 |g|^2 \\ &\geq n^2 \|g\|^2 \end{align*} which implies $$\|M_f\| \geq n$$ proving that $$M_f$$ is not bounded.

But I don't see how I can find such $$g.$$

P.S. If we assume that $$(\Omega,\mathcal{F},\mu)$$ is $$\sigma-$$finite, then I can find a subset of $$F_{n,m}$$ of $$E_n$$ of finite positive measure with $$n \leq |f|< n+m$$ on $$F_{n,m},$$ and let $$g=\mathbb{1}_{F_{n,m}}.$$

1 Answer

The claim is not true in general. If $$\mu$$ is such that $$\mu(A)=+\infty$$ for all $$A\ne\emptyset$$, then $$L^2=\{0\}$$ is trivial, and $$f$$ can be arbitrary.

If the measure space is such that every set $$A$$ of positive measure contains a subset of positive and finite measure then the claim is true. $$\sigma$$-finiteness is sufficient, as you wrote.

• Thanks for the example, @daw. I was wondering, what would be the necessary and sufficient condition for the claim to be true? Commented Jul 17, 2021 at 20:05
• Since I am only interested in the finite positive measures of $E_n,$ I think it is also sufficient that the sigma subalgebra generated by $f$ be sigma finite. Commented Jul 18, 2021 at 6:48