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Let $H$ be a Hilbert space and $T\in\mathcal{K}(H)$. Show that if $(x_n)_{n\in\mathbb{N}}$ is a sequence in $H$ that converges weakly to $x_0\in H$ then $\lim_{n\to\infty}||Tx_n-Tx_0||=0$.

My proof:

Since $\overline{T(B_1)}$ is compact and thus separable we conclude by scaling that $\overline{\text{ran}(T)}$ is a separable sub-Hilbert space of $H$ and thus has a countable (Schauder) basis $E:=\{e_i:i\in\mathbb{N}\}$.

Also we have $Tx=\sum_{i\in\mathbb{N}} \langle Tx,e_i\rangle e_i$ for $x\in H$.

Moreover a direct corollary from the Frèchet-Riesz theorem is that $x_n\overset{w}{\to}x_0$ $(n\to\infty)$ if and only if $\langle x_n-x_0,y\rangle\overset{n\to\infty}{\to} 0$ for all $y\in H$.

Also $T$ is bounded and therefore $T^*$ exists on all of $H$ and is bounded.

Now we bring everything together and have

$$\begin{aligned} ||Tx_n-Tx_0||^2 &=||\sum_{i\in\mathbb{N}}\langle T(x_n-x_0),e_i\rangle e_i||^2 &&=\sum_{i\in\mathbb{N}}|\langle T(x_n-x_0),e_i\rangle|^2\\ &= \sum_{i\in\mathbb{N}}|\langle (x_n-x_0),T^*e_i\rangle |^2 &&\overset{n\to\infty}{\to}0 \end{aligned} $$

My professor told me that I had to give another argument for the last step (taking the limit inside the sum). Can someone tell me, how I would argue?

mathemagician99
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  • FYI you seem to be assuming a Hilbert basis (an orthonormal Schauder basis), not merely a Schauder basis. There's nothing wrong with this, given we have a Hilbert space, but you should be aware that not all Schauder bases are quite so powerful. – user790072 May 25 '20 at 13:20
  • @user790072 Can you please give an argument why one can take the limit inside the sum? – mathemagician99 May 25 '20 at 17:53
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    It seems like a standard exercise, but I'm not particularly well-versed with compact operators. I've had a bit of a think about it, but I couldn't come up with an argument. I'll post an answer if I think of anything. – user790072 May 25 '20 at 18:36

1 Answers1

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A few comments:

  • A Schauder basis is not necessarily an orthonormal one, and so you wouldn't have the equality $Tx=\sum_j\langle Tx,e_j\rangle\,e_j$. You need an orthonormal basis.

  • You are right that it's a "direct corollary" of Riesz-Fréchet, but over a Hilbert space one just uses weak convergence as $\langle x_n-x,y\rangle\to0$.

  • As $x_n\to x$ if and only if $x_n-x\to0$, it is simpler (for writing) to assume that $x_n\to0$.

  • Note that nowhere in your argument have you used that $T$ is compact. The "taking the limit inside the sum" requires you to use that $T$ is compact.

  • And, also, the result is not true if you consider a weakly convergent net; it has to be a sequence. That's also something that your argument requires. Here, since $\{x_n\}$ is a weakly convergent sequence, it is bounded.

One possibility is to use that $T$, being compact, is a limit of finite-rank operators. If $T_0$ is finite-rank, then $T_0x=\sum_{j=1}^mf_j(x)\,g_j$ for certain bounded linear functionals $f_j$ and linearly independent vectors $g_1,\ldots,g_m$. So $$ \|T_0x_n\|\leq\sum_{j=1}^m|f_j(x_n)|\,\|g_j\|\to0 $$ since $f_j(x_n)\to0$ for all $j$. Then, for any $\varepsilon>0$ there exists $T_0$, finite-rank, with $\|T-T_0\|<\varepsilon$. Then $$ \|Tx_n\|\leq\|T_0x_n\|+\|(T-T_0)x_n\|\leq\|T_0x_n\|+\varepsilon\,\sup_n\|x_n\|. $$ Then $$ \limsup_n\|Tx_n\|\leq \varepsilon\,\sup_n\|x_n\|. $$ As we can do this for every $\varepsilon>0$, we have that the limit exists and is zero.

Martin Argerami
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  • Well, I actually used the compactness to argue that the closure of the range of $T$ is separable and thus has an orthonormal Schauder basis. Actually in my studies the terms orthonormal Schauder basis and just Schauder basis were used synonymous. Can you give the name or another source for the theorem that tells you that a finite rank operator can be written as above? – mathemagician99 May 25 '20 at 21:42
  • If $H$ is separable (which it usually is) then the range of any operator is separable, compact or not; so that cannot be enough for your proof. Also, usually you don't want to talk about a basis of the range of $T$, because it will be non-closed in general (a compact operator has closed range if and only if it has finite range). An orthonormal basis is automatically Schauder, so no one ever uses "Schauder". Finally, about the expression of the finite-rank operator: I'll be surprised if the proof is written in any book; I would expect it to be an (easy) exercise. – Martin Argerami May 25 '20 at 21:51
  • Well here don't have the condition that $H$ is separable and the point is that one has a countable orthonormal Schauder basis, so one does not have to worry about unconditional convergence. Regarding the statement about finite rank operators... if one knows that any operator with rank 1 is basically a functional times a vector then induction can be used. By why is that so? Why does an operator with rank 1 has to have that form? – mathemagician99 May 26 '20 at 09:52
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    But you don't need to use separability anywhere in the argument. My point was that you had only used compactness to say that your basis is countable, and so on a separable space your argument was supposedly showing that every operator satisfies the property (which is not true). As for finite-rank: if $\dim TH<\infty$, let $h_1,\ldots,h_n$ be an orthonormal basis of $TH$. Then $Tx=\sum_j\alpha_j(x),h_j$. Now show that each $\alpha_j$ is linear and bounded. – Martin Argerami May 26 '20 at 14:30