Adjoint operator of compact operator is compact

Asked May 25 '20 at 22:51

Active May 25 '20 at 22:51

Viewed 184 times

I have the following problem: Given $H$ a Hilbert space, and $T\in\mathcal{L}(H,H)$ a compact operator. Show that $T^*$ is compact, where $T^*$ is the adjoint of $T$. (That is the operator defined by $ \langle Tx,y\rangle = \langle x,T^*y \rangle $,$\forall x,y\in H$).

This is given in a class exercise where it is suggested that we use that $T^*T$ is compact. Currently I have absolutely no idea on how to use this.

There are other threads such as Easy Proof Adjoint(Compact)=Compact where a solution is given. However, they use heavy machinery currently out of reach, such as Banach-Alaouglu's or Arzeli-Ascoli's theorem. There must be a simple solution, anyone got a clue?

Thanks to everyone.

asked May 25 '20 at 22:51

miraunpajaro

1,560

Do you know that "an operator is compact iff it is the operator-limit of a sequence of finite-rank operators"? – Sahiba Arora May 25 '20 at 22:54
Also this answer doesn't seem to use any heavy machinery. – Sahiba Arora May 25 '20 at 22:55
@SahibaArora Only if $H$ is separable, which is not necessarily the case – miraunpajaro May 25 '20 at 23:07
@SahibaArora It uses that every bounded sequence has a weakly convergent subsequence, which is basically the Banach-Alaoglu's theorem – miraunpajaro May 25 '20 at 23:09

Adjoint operator of compact operator is compact

0 Answers0