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Let $\{\varphi_n\}$ be an orthonormal sequence in a Hilbert space $H$ and consider the operator $T:H \to H$ defined by

$$T(f) = \sum_{n=1}^\infty a_n \color{blue}{\langle \varphi_n, f \rangle} \varphi_n \tag{$\star$}$$

where $\{a_n\}$ is a sequence of scalars satisfying $\displaystyle\lim_{n \to \infty} a_n = 0$. Show that $T$ is a compact operator.

To show $T$ is compact, I mean that I must show that if $B$ is the unit ball via

$$B = \{f \in H: |f| \leq 1\}$$

then $T(B)$ is a compact set, i.e. that for every sequence $\{f_j\}_{k=1}^\infty \subseteq B$, there is a subsequence $\{f_{j_k}\}_{k=1}^\infty$ such that $\{T(f_{j_k})\}_{k=1}^\infty$ converges in the norm $\| \cdot \|$.

I'm assuming that the definition of $T$ in $(\star)$ carries the implicit assumption that the choice of $\{a_n\}_n$ depends upon $f_j$ since otherwise, if $\{a_n\}_n$ is a fixed sequence, then $\{T(f_j)\}_j$ is a constant sequence which trivializes the entire problem.

Let $\{f_j\}_j \subseteq B$ be a bounded sequence, i.e. $\exists \, M > 0$ s.t. $\|f_j\| \leq M$. Now, upon examination, for each $j$,

$$T(f_j) = \sum_{n=1}^\infty a_n^{(j)}\color{blue}{\langle \varphi_n, f \rangle} \varphi_n \tag{$\star\star$}$$

It's not clear what I should do here. I suspect that my assumption that $\{a_n\}_n$ depends on $f_j$ is incorrect given the wording of the problem. But if $\{a_n\}_n$ is fixed, then $T$ is a constant map, which makes this problem silly. Maybe the way I'm interpreting this problem is incorrect, but I cannot see it any other way. Am I understanding the problem correctly?

EDIT: The text in blue in $(\star)$ is a restatement of the problem after Anne's comment mentioned that there may be a typo in the original problem $-$ which originally omitted the blue text in $(\star)$.

Attempt:

Note that operators $S$ of finite rank ($\dim(S(H)) < \infty$) are compact. If $T$ is the limit of a sequence of finite rank (hence, compact) operators, then $T$ is compact. Define $T_m: H \to H$ via

\begin{align} T_m(f) = \sum_{i=1}^m a_n \langle \varphi_n,f \rangle \varphi_n \end{align}

If we take $(T - T_m)(f) = T(f) - T_m(f)$, we get

\begin{align} \begin{split} T(f) -T_m(f) &= \sum_{n=1}^\infty a_n \langle \varphi_n, f \rangle \varphi_n - \sum_{n=1}^m a_n \langle \varphi_n, f \rangle \varphi_n \\ &= \sum_{n=m+1}^\infty a_n \langle \varphi_n ,f \rangle \varphi_n \end{split} \end{align}

And now consider $\|(T-T_m)(f)\|^2$:

\begin{align} \begin{split} \|(T-T_m)(f)\|^2 &= \left\| \sum_{n=m+1}^\infty a_n \langle \varphi_n ,f \rangle \varphi_n \right\|^2 \\ &= \left\langle \sum_{n=m+1}^\infty a_n \langle \varphi_n ,f \rangle \varphi_n, \sum_{n=m+1}^\infty a_n \langle \varphi_n ,f \rangle \varphi_n \right\rangle \\ &= \sum_{n=m+1}^\infty \left\langle a_n \langle \varphi_n ,f \rangle \varphi_n, a_n \langle \varphi_n ,f \rangle \varphi_n \right\rangle \\ &= \sum_{n=m+1}^\infty \| a_n \langle \varphi_n ,f \rangle \varphi_n \|^2 \\ &= \sum_{n=m+1}^\infty |a_n|^{\color{red}2} \cdot \| \langle \varphi_n ,f \rangle \varphi_n \|^2 \\ \end{split} \end{align}

Aram Nazaryan
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    There must be a typo in the definition of $T$. I guess $(a_n)$ is a fixed sequence converging to $0$, and $T(f)=\sum a_n\langle\varphi_n,f\rangle$. And to prove that $T$ is compact, you can check that it is a limit (for the operator norm) of finite-rank operators: $T_m:=\sum_{n\le m} a_n\langle\varphi_n,f\rangle$. – Anne Bauval May 20 '24 at 21:48
  • Oops my formulas for $T$ and $T_m$ were not correct. Replace $a_n\langle\varphi_n,f\rangle$ with $a_n\langle\varphi_n,f\rangle\color{red}{\varphi_n}$. Or equivalently: $a_nf_n\varphi_n$, where $f_n:=\langle\varphi_n,f\rangle$. – Anne Bauval May 20 '24 at 21:59
  • @AnneBauval I can see how the extra $\varphi_n$ was necessary to add in your edit (since $\varphi_n$ is in the definition of $T$). But what is the connection between the product $a_n \varphi_n$ in the definition of $T$ and the product $a_n \langle \varphi_n, f \rangle \varphi_n$ in the definition of $T_m$? i.e. what is the intuition behind incorporating the $\langle \varphi_n, f \rangle$ into $T_m$ - as it does not appear in $T$? – Aram Nazaryan May 20 '24 at 22:05
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    I inserted $\langle\varphi_n,f\rangle$ in the definition of $T$ as well. That is why I said there must be a typo. This reminds me of https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Spectral_theorem – Anne Bauval May 20 '24 at 22:10
  • @AnneBauval I see that $T-T_m$ is the tail of $T$ from the $m+1^{th}$ summand onward. So I will be summing $a_j \langle \varphi_j, f \rangle \varphi_n$ as $j \geq m+1$. Now, $a_j \to 0$ doesn't necessarily imply that $\sum a_j \to 0$ so it's hard to see how $a_j \to 0$ is necessary here. I'm struggling to use the orthonormality of $\varphi_n$ as well since we're not given an orthonormal basis with which we can rewrite $f$ (unless every infinite orthonormal subset of an infinite dimensional Hilbert space is a basis as well) – Aram Nazaryan May 20 '24 at 22:21
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    @AnneBauval I have posted an attempt at the bottom of my post – Aram Nazaryan May 21 '24 at 04:10
  • There is an easier and slightly more general approach based on the definition https://math.stackexchange.com/a/416111/27978 – copper.hat May 21 '24 at 05:18
  • Related to this. – jd27 May 21 '24 at 07:01
  • Does this answer your question? Showing that if $\lim_{i\to\infty}\lambda_i=0$ and $Ae_i=\lambda_ie_i$ for $A\in\mathcal{B}(H)$, then $A$ is a compact operator Actually, it does (the fact that your ${\varphi_n}$ is only orthonormal and not a Hilbert basis makes no difference). @jd27 Good catch! (but posting an answer just after providing that link is redundant). – Anne Bauval May 21 '24 at 09:16
  • @copper.hat easier and more general than what? Proving that $T$ is a limit of finite rank operators is precisely the advice (mainly in my first comment) that Grigor almost finished to achieve (in his last edit 1 hour before your comment). – Anne Bauval May 21 '24 at 09:26
  • Grigor, you are almost done (you forgot a square on your last $|a_n|$). You can conclude that $$ |(T-T_m)(f)|^2=\sum_{n>m}|a_n|^2|\langle\varphi_n,f\rangle|^2\le(\sup_{n>m}|a_n|)^2|f|^2$$ (as written in my message to you 11 hours ago, which I shall now delete) and therefore, $|(T-T_m)|\le\sup_{n>m}|a_n|\to0$, and you are done. – Anne Bauval May 21 '24 at 09:37
  • One small remark about your 'attempt': since you insist on writing the sequence of inequalities in full detail, you could add an intermediate step: $$\dots=\sum_{n=m+1}^\infty\sum_{k=m+1}^\infty\left\langle a_n \langle \varphi_n ,f \rangle \varphi_n, a_k\langle \varphi_k,f \rangle \varphi_k\right\rangle =\dots$$ – Anne Bauval May 21 '24 at 09:38
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    @AnneBauval I think that OP's question is sufficiently different to the linked one, since OP is asking about a specific step in the proof (which is neither asked about nor answered in the linked post). – jd27 May 21 '24 at 10:18
  • @AnneBauval My apologies, my intent was not to offend. – copper.hat May 21 '24 at 18:11

1 Answers1

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Let $f \in H$ arbitrary. Starting from your final sequence of equalities (you forgot a square): $$ \begin{align} \begin{split} \|(T-T_m)(f)\|^2 &= \sum_{n=m+1}^\infty |a_n|^2 \cdot \| \langle \varphi_n ,f \rangle \varphi_n \|^2 \\ & \leq (\sup_{n \in \mathbb{N}: n \geq m+1} |a_n|)^2 \sum_{n=m+1}^\infty \| \langle \varphi_n ,f \rangle \varphi_n \|^2 \\ &\leq (\sup_{n \in \mathbb{N}: n \geq m+1} |a_n|)^2 \| f\|^2. \end{split} \end{align} $$ Since $f$ is arbitrary this implies that $$ 0 \leq \| T- T_m \| \leq \sup_{n \in \mathbb{N}: n \geq m+1} |a_n| $$ for all $m \in \mathbb{N}$. The sequence on the right hand side converges to $0$ (as $m \to \infty$), because it converges to the $ \limsup $ of $ (|a_n|)_{n \in \mathbb{N}} $, which is $0$. Therefore the Sandwich lemma concludes.

jd27
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  • Thank you for explaining this so well. Everything looks good - except I'm wondering about the justification behind $\sum_{n =m+1}^\infty | \langle \varphi_n, f \rangle \varphi_n |^2 \leq |f|^2$, which is implict from your second $\leq$. Could you elaborate on this part? – Aram Nazaryan May 21 '24 at 13:17
  • @GrigorHakobyan $| \langle \varphi_n , f \rangle \varphi |^2 = |\langle \varphi_n, f \rangle |^2$ for any $n \in \mathbb{N}$ and then the inequality is just Bessels inequality. – jd27 May 21 '24 at 13:51
  • Ah, Bessel's inequality makes sense, but I'm still having trouble seeing $| \langle \varphi_n , f \rangle \varphi_n |^2 = | \langle \varphi_n ,f \rangle |^2$. I know that $| \xi |^2 = \langle \xi, \xi \rangle$ in general. Letting $\xi = \langle \varphi_n, f \rangle \varphi_n$, then we have $|\langle \varphi_n, f \rangle \varphi_n|^2 = \langle \langle \varphi_n, f \rangle \varphi_n, \langle \varphi_n, f \rangle \varphi_n \rangle$? Still not sure how you obtain $| \langle \varphi_n , f \rangle \varphi_n |^2 = | \langle \varphi_n ,f \rangle |^2$ – Aram Nazaryan May 21 '24 at 14:05
  • @GrigorHakobyan $| \langle \varphi_n, f \rangle \varphi_n |^2 = |\langle \varphi_n, f \rangle|^2 | \varphi_n |^2$ by the absolute homogenity of the norm and $| \varphi_n |=1$ (since $\varphi_n$ is an element of an orthonormal set). – jd27 May 21 '24 at 14:16
  • @Grigor You also used tacitely that $\forall k\ne n\quad\langle\varphi_n,\varphi_k\rangle=0$ (see my comment above beginning with 'One small remark'). – Anne Bauval May 21 '24 at 15:01