Prove an operator is compact and classify the spectral values Let $T(\eta) = \{\frac{1}{2^n}\eta_n\}$ for $\eta = \{\eta_1, \eta_2, ...\} \in \ell_2$. I need to prove T is compact, and evaluate $\sigma_p(T), \sigma_c(T), \sigma_r(T)$. 
My feeling is that I need to show that every bounded sequences of $\ell_2$ sequence $x_n$ will contain a convergent subsequence in $T(x_n)$, with some less than inequality I should be able to show that, but I failed to reach such property. 
 A: Let $\{ e_{n} \}_{n=1}^{\infty}$ be the standard basis for $\ell^{2}$. Specifically,
$$
              e_1 = \{ 1,0,0,0,\cdots \} \\
              e_2 = \{ 0,1,0,0,\cdots \} \\
              e_3 = \{ 0,0,1,0,\cdots \} \\
                     \vdots
$$
You can write
$$
                 Tx = \sum_{n=1}^{\infty}\frac{1}{2^{n}}(x,e_n)e_n
$$
If you truncate this series to $n \le N$, then
\begin{align}
    \left\|Tx - \sum_{n=1}^{N}\frac{1}{2^{n}}(x,e_n)e_n\right\|^{2}
              & = \sum_{n=N+1}^{\infty}\frac{1}{2^{2n}}|(x,e_n)|^{2} \\
              & \le  \frac{1}{2^{2N+2}}\sum_{n=N+1}|(x,e_n)|^{2} \\
              & \le  \frac{1}{2^{2N+2}}\sum_{n=1}^{\infty}|(x,e_n)|^{2} \\
              &  =   \frac{1}{2^{2N+2}}\|x\|^{2}.
\end{align}
That means the finite-rank operator $T_{N}x=\sum_{n=1}^{N}\frac{1}{2^{n}}(x,e_n)e_n$ satisfies
$$
                     \|Tx-T_{N}x \| \le \frac{1}{2^{N+1}}\|x\| \\
               \implies \|T-T_{N}\|_{\mathcal{L}(X)} \le \frac{1}{2^{N+1}}\rightarrow 0.
$$
You can see that $1/2^{n}$ is an eigenvalue because $Te_{n}=\frac{1}{2^{n}}e_{n}$. Compact operators can only have point spectrum for the non-zero spectrum. $0$ is a cluster point of the spectrum in this case, and $T$ is not continuously invertible even though it is injective.
A: Please note that if $T$ were surjective, then its inverse must be continuous (if $T$ is compact, it is closed, hence so must be $T^{-1}$ if it exists. By the closed graph theorem, $T^{-1}$ is then continuous).
As T.A.E. explained, $\{1/2^n :n=1,2,3,\dots\}\subseteq \sigma_p(T)$.
It is easy to see that no other numbers belong to the point spectrum of $T$.
This already shows that
$\sigma(T)\setminus\{0\} = \sigma_p(T)\setminus\{0\} = \{1/2^n :n=1,2,3,\dots\}$.
Since $\sigma(T)$ is closed, $0$ must belong to the spectrum. Hence $T$ cannot be surjective. There are (at least) two possibilities to see that $0\in\sigma_c(T)$:
1) Clearly $T$ in injective and all vectors of the standard orthonormal basis of $\ell_2$ belong to the range of $T$. Hence the range of $T$ is dense in $\ell_2$ and therefore $0\in\sigma_c(T)$.
2) Clearly $T$ is selfadjoint. Hence $\sigma_r(T)=\emptyset$. Since $0\in\sigma(T)$ and $0\notin\sigma_p(T)\cup\sigma_r(T)$, the only remaining possibility is $0\in\sigma_c(T)$.
