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Hell0 there!

I have to show whether the operator $$ T\colon L^2(\mathbb{R})\to L^2(\mathbb{R}), f\mapsto\chi_{[0,1]}f $$ is continuous, selfadjoint and compact. I have problems to show the compactness.

But first let me show my proofs concerning the other two points:

1.) $T$ is indeed continuous, because it is bounded:

$$ \lVert Tf\rVert_{L^2}=\int\limits_{\mathbb{R}}\lvert\chi_{[0,1]}(x)f(x)\rvert^2\, dx=\int\limits_0^1\lvert f(x)\rvert^2\, dx\leq\int\limits_{\mathbb{R}}\lvert f(x)\rvert^2\, dx=\lVert f\rVert_{L^2} $$

2.) $T$ is selfadjoint:

$$ \langle Tf,g\rangle_{L^2}=\int\limits_{\mathbb{R}}f(x)\chi_{[0,1]}(x)\overline{g(x)}\, dx=\int\limits_{\mathbb{R}}f(x)\overline{\chi_{[0,1]}(x)g(x)}\, dx=\langle f,T^*g\rangle_{L^2} $$

and therefore the adjoint operator is given by $f\mapsto\chi_{[0,1]}f$ which is $T$ itself.

3.) Compactness:

Now I do not know how to show if $T$ is compact or not. There are two different criteria of compactness that would be useful here (to my opinion):

(a) Show that the unit ball is compact relatively.

(b) Show that for every bounded sequence $(f_n)$ the sequence $(Tf_n)$ has a convergent subsequence.

Which is the best to use here?

Greetings & have a good time


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In 3.)(a), do you mean the image of the unit ball? – Jonas Meyer Feb 6 '13 at 17:04
Yes, excuse me. I mean the image of the unit ball. – math12 Feb 6 '13 at 17:06
up vote 2 down vote accepted

Consider the subspace $H\subseteq L^2$ defined by $H=\{f\in L^2: Tf=f\}\cong L^2[0,1]$. You can show that $T|_H$ is not compact as an operator on $H$, and that a restriction of a compact operator is compact. (Or you can argue directly by choosing your sequence referred to in 3.)(b) from $H$.) Infinite dimensional normed spaces like $H$ always have bounded sequences with no convergent subsequences, and in this case you can find some fairly simple concrete examples.

In general, the eigenspaces of a compact operator corresponding to nonzero eigenvalues must be finite dimensional, because the identity operator on an infinite dimensional normed space is not compact.

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I think $H$ has got an orthonormal basis $\left\{e_n\right\}$, because it is a Hilbertspace itself. I think, then $(e_n)$ is a bounded sequence and $\lVert Te_n-Te_m\rVert=\lVert e_n-e_m\rVert=\sqrt{2}$ and therefore $(Te_n)$ cannot have a convergent subsequence. Is that right? – math12 Feb 6 '13 at 17:30
@math12: Yes. And if you want to be even more explicit, such $(e_n)$ can be given by trig functions. But every infinite linearly independent sequence in $L^2[0,1]$ (e.g., $(1,x,x^2,\ldots)$) can be Gram-Schmidt'ed to an orthonormal sequence. – Jonas Meyer Feb 6 '13 at 17:33

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