A not compact operator defined in $l^2$

Question

I'm trying to solve the following problem presented in Kreyszig's Introductory Functional Analysis with Applications. Consider the operator $T:l^2\to l^2$ defined by $$Tx = (\alpha_k\xi_k), \quad x=(\xi_k)\in l^2,$$ where the sequence $(\alpha_k)$ is dense in $[0,1]$. I have to prove that $T$ is not compact, so I consider the bounded sequence $(e_n)$ given by $$e_n = (0,0,\ldots,1_n,0,0,\ldots).$$ We then obtain the sequence $(Te_n)$ given by $$Te_n = (0,0,\ldots,\alpha_n,0,0,\ldots).$$

I don't know how to justify that $(Te_n)$ doesn't have a convergent subsequence, or even worse if this sequence is the appropiate to proceed. In the solution given, the author says "every $\alpha_k$ is an eigenvalue of $T$" (so you can apply a result that every sequence of eigenvalues of a compact operators must converge to zero and obtain a contradiction) but that clearly is not correct.

I would appreciate if someone can help me with this.

Answer 1

One easy way of proving this is to use the theorem which says that any compact operator on a Hilbert space has at most countbaly many eigen values which, if infinite in number, form a sequence tending to $ 0$. Note that $Te_n=\alpha_n e_n$ so each $\alpha_i$ is an eigen value. But if the $\alpha_i$'s form a sequence tending to $0$ they cannot be dense in $[0,1]$.

Answer 2

You could also proceed with the same idea in your original proof without referring to the spectral theory of compact operators:

Since $\alpha_n$ is bounded and not tending to zero, there is a subsequence $n_k$ with $\alpha_{n_k}\to c\ne0$. Now the sequence $e_{n_k}$ is bounded, and its image sequence $Te_{n_k}$ has no convergent subsequence. Indeed, if it would have a convergent subsequence $Te_{n_{k_\ell}}\to y$ then its limit $y$ would have to be $0$ (why?), but $\alpha_{n_{k_\ell}}\to c\ne0$, a contradiction (why?).