# Multiplication Operator on $L^2$ is densely defined

This is an exercise in Terence Tao's notes on spectral theory

Let $$(X, \mu)$$ be a measure space with a countable generated $$\sigma$$-algebra and $$m: X \to \mathbb{R}$$ a measurable function. Let $$D$$ be the space of all $$f \in L^2(X)$$ such that $$mf \in L^2(X)$$. Show that the operator $$L: D \to L^2$$ defined by $$Lf:= mf$$ is a densely defined self-adjoint operator.

The self-adjoint part is fine, and it is clear to me that if $$m$$ were bounded then I can just see that simple functions belong to $$D$$ making it densely defined. However, I feel like since $$m$$ is just measurable it's possible to make it very non-integrable on lots of sets so that $$D$$ might not be dense. Can anyone give me an idea for why $$D$$ must be a dense subset of $$L^2$$?

• A self-adjoint operator is always densely defined. I guess you have just shown symmetry of the operator. In fact, I asked a very similar question just a few days ago: math.stackexchange.com/questions/3162251/… Robert Israel showed very nicely that the operator is densely defined. However, the adjoint part is still missing. – amsmath Mar 28 '19 at 18:18

Suppose $$f \perp\mathcal{D}(L)$$. Then $$\frac{1}{m^2+1}f\in\mathcal{D}(L)$$, which forces $$f \perp \frac{1}{m^2+1}f$$ and leads to $$\frac{1}{m^2+1}|f|^2 = 0$$ a.e.. Hence, $$f=0$$ a.e.. So $$\mathcal{D}(L)$$ is dense in $$L^2(X,\mu)$$.