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}
Best Answer
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.