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I have the following problem:

Let H be a Hilbert space

a) Prove that if $T: H\to H$ is compact then $T^2$ is compact operator

b) Find $S: H\to H$ compact such that $S=T^2$ with T non compact

c)If T is self adjoint then $T^2$ compact implies T is compact.

I have managed to prove (a). Can anyone give me any idea for the other requests. Thank you in advance.

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  • $\begingroup$ b) Let $H_1$ be closed infinite dimensional subspace of $H$, such that $H_1$ and $H_2:=H_1^\perp$ are of equal Hilbert dimension. Show that $$ T:H_1\oplus_2 H_2\to H_1\oplus_2 H_2: x_1\oplus_2 x_2\mapsto x_2\oplus_2 0 $$ is the desired operator $\endgroup$ – Norbert Dec 3 '13 at 0:11
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a) You use that the compacts are an ideal.

b) Let $T$ be the operator defined on the canonical basis by $Te_{2j-1}=0$, $Te_{2j}=e_{2j+1}$, $j\in\mathbb N$. Then $T$ is not compact, as it maps an infinite orthonormal set into another. But $T^2=0$, which of course is compact.

c) For any $T$, if $T^*T$ is compact, then so is $T$. Because then $|T|=(T^*T)^{1/2}$ is compact (as it is a limit of compacts), and $T=V|T|$ by the polar decomposition. This implies the assertion in the selfadjoint case.

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  • $\begingroup$ In this answer you are assuming $H$ is separable, though it was not given. By the way is there a simpler argument for c). I had the sme soltion but it seems for me as mosquito nuking. $\endgroup$ – Norbert Dec 3 '13 at 12:53
  • $\begingroup$ Yes and no. It is mostly a matter of notation. One can make the exact same argument by replacing "the canonical basis" with "an orthonormal set $\{e_j\}_{j\in\mathbb N}$". But it looks to me that it would only make it a little harder to follow for those with little experience. Regarding c), all I could think of was this argument, and one using spectral projections, which is even more technical than the one I posted. $\endgroup$ – Martin Argerami Dec 3 '13 at 15:48
  • $\begingroup$ Is $S$ as defined in (b) self-adjoint? What is $S^{*}$? Because if that were the case, then $S$ would in fact be normal. $\endgroup$ – Libertron Dec 8 '13 at 23:30
  • $\begingroup$ No, of course $S$ is not self-adjoint. It is not even normal. For a normal, $S^2=0$ implies $S=0$. That said, I should have respected the notation in the question and use $T$ instead of $S$. I'll edit. $\endgroup$ – Martin Argerami Dec 9 '13 at 0:42

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