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Let $H$ be a Hilbert space and $(T_n)_{n \in \mathbb{N}}$ be a sequence of bounded linear operators on $H$.

The strong convergence of $T_n$ doesn't imply the strong convergence of $T_n^*$, i.e. $\|T_n(x)\| \rightarrow 0 ~\forall x \in H$ doesn't imply that $\|T_n^*(x)\| \rightarrow 0 ~\forall x \in H$.

The easiest counterexemple that I can think of is $T_n := (S^*)^n$ where $S$ is the right-shift on $\ell^2$ as $\|T_n(x)\| \rightarrow 0 ~\forall x \in H$ but $\|T_n^*(x)\| = \|S^n(x)\| = \|x\|$ if $x \neq 0$.

Does the additiinnal hypothesis of $(T_n)_{n \in \mathbb{N}}$ being a sequence of compact operator change the result ?

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up vote 2 down vote accepted

I don't think your additional hypothesis will make this true.

Let $S$ be the rightward shift of $\ell^2(\mathbb{N})$ and let $D e_i = e_i /i$, where $\{e_i\}$ is the standard basis of $\ell^2$. Note that $D$ is compact and self-adjoint.

Take $T_n = D \circ (S^*)^n$, which are all compact, so that like before, $T_n \rightarrow 0$ strongly. On the other hand, $T_n^* = S^n \circ D$ and $$\|T_n^* x\| = \|D x\|$$ for all $x \in \ell^2$, $n \in \mathbb{N}$. The same recipe holds for any self-adjoint, compact operator $D$.

I interpret this in the following way: there are stronger statements to make when you enforce compactness of the sequence $\{T_n\}$, but somehow the compact images of the unit ball $B$ under $T_n$ may be moved around enough to avoid each other.

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Thanks ! This interpretation is quite enlightening. – M.G Apr 9 '13 at 0:36

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