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Let $T: \mathcal S(\mathbb R)\to L^2(\mathbb R)$ defined by $$Tf(x)=f'(x),$$ where $\mathcal S(\mathbb R)$ is the Schwarz space on $\mathbb R$. The question is : is there a continuous extension to $L^2(\mathbb R)$, and the answer is no, because $\frac{\|Tf_k\|}{\|f_k\|}\to \infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.

You'll see the example here

1) Why is this a counter-example since $e^{-k|x|}\notin \mathcal S(\mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $\mathbb R$, is it ? Because on $L^2(\mathbb R)\setminus \mathcal S(\mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $\mathcal S(\mathbb R)$ than on $L^2(\mathbb R)$.

2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).

user621345
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  • If $f$ is derivable a.e. and $f'\in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set. – Surb Jan 06 '19 at 16:07

1 Answers1

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By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous. So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.

Robert Israel
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    So if $T: L^2\to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where... – user621345 Jan 06 '19 at 16:12
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    It is consistent with ZF+DC that every $T: L^2 \to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded. – Robert Israel Jan 06 '19 at 19:50