# functional analysis – 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 $$gin L^2(Omega,mathcal{F},mu)$$ 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 $$E_n={w inOmega: |f(w)| geq n}$$ 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)|&= 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 $$g=mathbb{1}_{F_{n,m}}.$$

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