Let $H$ be a Hilbert space and $\{\phi_k\}$ an orthonormal basis of $H$. Prove that the linear operator $T:H \to H$ defined by $T(\phi_k) = \frac{1}{k+1}\phi_{k+1}$ is compact.
Proof: Consider the sequence of linear operators defined by $$ T_n(f) = \sum_{k=1}^{n}\frac{1}{k+1}\langle f,\phi_{k}\rangle\phi_{k+1}.$$ They have finite rank, so they are obviously all compact.
Since $T$ is linear and $\phi_k$ is an orthonormal basis, $$ T(f)=T(\sum_{k=1}^{\infty}\langle f, \phi_k \rangle\phi_k) =\sum_{k=1}^{\infty}\langle f, \phi_k \rangle T(\phi_k) = \sum_{k=1}^{\infty}\frac{1}{k+1}\langle f, \phi_k\rangle \phi_{k+1}.$$
Then $$ \|(T-T_n)(f)\|^2 = \|\sum_{k=n+1}^{\infty}\frac{1}{k+1}\langle f,\phi_k \rangle \phi_{k+1}\|^2 = \sum_{k=n+1}^{\infty}\frac{|\langle f,\phi_k\rangle |^2}{(k+1)^2}\leq\frac{1}{(n+2)^2}\sum_{k=n+1}^{\infty}|\langle f,\phi_{k}\rangle|^2 \leq \frac{1}{(n+2)^2}\sum_{k=1}^{\infty}|\langle f,\phi_{k}\rangle|^2 = \frac{1}{(n+2)^2}\|f\|^2.$$
So $$ \|T-T_n\|=\sup_{\|f\|=1}\|(T-T_n)(f)\| \leq \frac{1}{(n+2)^2},$$ which tends to zero as $n \to \infty$.
My question is: Why does this prove what was asked? This proves that there is a sequence of compact operators that converges to $T$. How can we infer that $T$ is compact from that? Is there such a theorem?
