Equivalent condition to an operator $T$ on a complex Hilbert space being compact

Question

Let $H$ be a complex Hilbert space and let $T:H\rightarrow H$ be a bounded linear map. The problem asks to show

$T$ is compact if and only if for any sequence $x_n$ weakly converging to $0$, $\langle Tx_n,x_n \rangle\rightarrow 0$.

While the forward implication seem to be relatively straightforward, I'm having trouble proving the reverse implication. By reflexivity of $H$ it suffices to show that $\langle Tx_n,x_n\rangle \rightarrow 0$ implies $\|Tx_n\|\rightarrow 0$. Also because the problem sort of suggests that something might be different for real Hilbert spaces, I tried using the polarization identity (but couldn't make it work). Any help would be appreciated. Thanks in advance!

Answer 1

As pointed out in the comments, the fact that the Hilbert space is complex plays a crucial role: if $H$ is the space of real square summable sequences and $T$ is such that $T(e_{2k})=e_{2k+1}$ and $T(e_{2k+1})=-e_{2k}$, where $e_j$ is the element of $H$ whose coordinate $j$ is one and all the others are zero, then $\langle Tx,x\rangle=0$ is satisfied for all $x\in H$ but $T$ is not compact because $e_{2k}\to 0$ weakly but $(T(e_{2k}))_{k\geqslant 1}$ does not admit a strongly convergent sequence.

Suppose that $H$ is a complex Hilbert space and $T\colon H\to H$ is linear, bounded and such that $\langle Tz_n,z_n\rangle \to 0$ for each sequence $(z_n)$ weakly convergent to $0$ and let us show that $T$ is compact.

We first show that $\langle Tx_n,y_n\rangle \to 0$ for each sequences $(x_n)$ and $(y_n)$ weakly convergent to $0$. Let $(x_n)$ and $(y_n)$ be such sequences. By looking at $\langle T(x_n+iy_n),x_n+iy_n\rangle$, we can see that $\langle Tx_n,y_n\rangle-\langle Ty_n,x_n\rangle\to 0$. This is not yet enough to conclude. But looking at $\langle T(x_n+ y_n),x_n+ y_n\rangle$, we can see that $\langle Tx_n,y_n\rangle+\langle Ty_n,x_n\rangle\to 0$ hence $\langle Tx_n,y_n\rangle \to 0$.

Now, in order to show that $T$ is compact, let $(x_n)$ be a sequence which converges weakly to $0$ and let us show that $\lVert Tx_{k}\rVert \to 0$. We know that $Tx_k\to 0$ weakly in $H$ hence letting $y_k=Tx_k$ and applying the previous fact gives that $\langle Tx_k,Tx_k\rangle\to 0$, which concludes the proof.

Answer 2

In fact, use spectra decomposition, the following stronger result is also true (for complex case): $T$ is compact if $\langle Te_n, e_n \rangle$ tend to zero for every orthonormal basis $\{e_n\}$, then $T$ is compact.

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For the proof, use the fact $$T=(T+T^*)/2+i(T-T^*)/2i,$$ we can assume $T$ is self-adjoint. Then suppose $$T=\int_{\mathbb{R}}x dE$$ is the spectra decomposition of $T$. Argue by contradiction, assume $T$ is not compact. We claim: there exists $\epsilon > 0$, such that either projection operator $E(-\infty,-\epsilon)$ or $E(\epsilon,+\infty)$ is infinite dimensional. The proof of the claim is again argue by contradiction and approach T by finite-rank operator, which can be found in Conway’s functional analysis Chapter IX, proposition 4.1. Now set $\epsilon > 0$ as above. We may assume the range of $E(\epsilon,+\infty)$ is infinite dimensional. Choose a orthonormal basis of ${\rm Ran} E$, say $\{ e_n\}$. Use functional calculation, one can check $\langle Te_n,e_n\rangle \geq \epsilon$. Extend $\{e_n\}$ to be a basis of whole $\mathcal{H}$, we get a contradiction.