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(I don't have a direct reference so this is a little fishy and I'll delete it if nobody recognises what I'm talking about, but I though for starters I'll ask anyway)

I've heard at university that if you have a operator (linear Hilbert space operator?) which is bounded on a restricted (compact?) domain, then it's bounded on the whole space. Put differently, to prove that an operator is bounded, you don't actually need to show it in the whole space.

However, if you restrict the domain of the position operator it becomes unbouned right? That's a contradiction.

In a finite dimensional example I see that you just have to check eigenvalues and hence consider the action on the base. The domain of the position operator is the function space with deltas as its base right? does restricting the domain here mean cut of $\delta(x-a)$ after $a\in \mathbb{R}$ has reached a certain value? I.e. considering only functions of compact support?

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1 Answer 1

In finite dimensions all operators are bounded so that case is not really relevant.

In the case of the position operator, you can't have it act on delta functions since these are not normalisable, despite how convenient this is for certain calculations. And it's unbounded anyway: consider it as an operator defined on some subspace of $L^2(\mathbb{R})$, then have it act on $\frac{1}{\sqrt{x^2+1}}$. This takes an element of finite norm to one of infinite norm hence the position operator is unbounded.

As for what you heard at university, I initially thought you could do that, but then I checked at MathOverflow and there is a simple counterexample to that claim, even if you take "restricted" domain to mean "dense".

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