Let $X$ be a Hilbert space having a countable orthonormal base $[u_1, u_2, \cdots]$. Also, suppose $Ax = \sum _{n=1} ^{\infty} \alpha_n \langle x,u_n \rangle u_n $, where $[\alpha_n]$ is a real bounded sequence. Is $A$ compact?
Now, I am familiar with the Spectral theorem, but that theorem starts with assuming that $A$ is indeed compact and also $\lim_{n \rightarrow \infty} \alpha_n=0$. Here, we only know that the sequence $[\alpha_n]$ is bounded.
The problem, as described above, resembles yet another unanswered question here: Show if $\lim_{n \to \infty} \lambda_n=0$ then $Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$ defines a compact operator.
I began with the very definition of a compact operator: considered a bounded sequence $[x_n]$ in $X$ and tried to show that $Tx_n$ has a convergent subsequence. But that led to a very general sequence. Is this the right approach to the problem? I guess there must be an easy solution to this one as it almost entirely resembles the spectral theorem. I'd appreciate any hint/help.
In general A is not compact, take for example A = I, and $\alpha_n = 1$. The confusion is in your notation, that is $$Ax = \sum _{n=1} ^{\infty} \alpha_n \langle x,u_n \rangle u_n$$. What does "=" mean here? If it meant norm convergence then A would be a limit of finite rank (hence compact) operators so it would be compact (this happens when $\alpha_n$ tends to $0$). But, I think the meaning of "=" is pointwise, i.e. for each $x$ the value of $Ax$ is the limit of the series. As remarked at the top this does not imply that $A$ is compact.