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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 $L^2$?

Memeboy Inc.
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  • 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: https://math.stackexchange.com/questions/3162251/adjoint-of-multiplication-operator 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
  • Where exactly does the property that the $\sigma$-algebra is countably generated come into the proof? – Inzinity Mar 29 '25 at 16:51

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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 $L^2(X,\mu)$.

amsmath
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Disintegrating By Parts
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