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Can somebody give me an easy example of a linear operator which maps $L^1(\mathbb{R}^n)$ to $L^1(\mathbb{R}^n)$ and $L^\infty(\mathbb{R}^n)$ to $L^\infty(\mathbb{R}^n)$ (but not boundedly) but does not admit a bounded extension from $L^2(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$ (or any other $L^p$, $1<p<\infty$) ?

Julien
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Anonymous999
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    You've done a search on google? – Elias Costa Jun 01 '13 at 11:35
  • What about a differentiation operator, say w.r.t. one coordinate? – Berci Jun 01 '13 at 11:45
  • And how should I extend it to L^1 and L^inf? – Anonymous999 Jun 01 '13 at 12:20
  • @Berci Could you precise what you have in mind? I understand you suggest $f\mapsto f'$ possibly extended by density. But this does not even send the bounded differentiable functions to $L^\infty$ ($x^2\sin (1/x^2)$). – Julien Jun 02 '13 at 01:44
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    @Anonymous999 You can extend linear operators using some (non-explicit) algebraic basis of $L^1$ or $L^\infty$. Follow the proof of the Hahn-Banach theorem, but drop the majorizing sublinear functional from it. – ˈjuː.zɚ79365 Jun 02 '13 at 12:10

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There are no explicit (easy or otherwise) examples of unbounded linear operators (or functionals) defined on a Banach space. Their very existence depends on the axiom of choice. See Discontinuous linear functional.

Community
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ˈjuː.zɚ79365
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  • That's only if you require your operator to be defined everywhere. Otherwise it is very easy, like differentiation on $C^1[0,1]$ as a dense subspace of $C[0,1]$. In ththe case of the question above, it is clearly not asking for an everywhere-defined operator. – Martin Argerami Jun 02 '13 at 05:15
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    @MartinArgerami Clearly? The question says "maps $L^1(\mathbb R^n)$..", no mention of a dense subspace. When suggested a densely defined operator (differentiation) OP responded with "And how should I extend it to $L^1$ and $L^\infty$?" – ˈjuː.zɚ79365 Jun 02 '13 at 06:32
  • Well, I have to admit that not "clearly". Since none of the spaces the OP mentions are included in each other, I'm not sure what he is looking for. – Martin Argerami Jun 02 '13 at 13:30