Let $(a_n)$ be a sequence that tends to zero and $0
Someone said I should look at its inverse, but that didn't helpt me either...
Let $(a_n)$ be a sequence that tends to zero and $0
Someone said I should look at its inverse, but that didn't helpt me either...
I think looking at bounded sequence $e_n=(0, \ldots ,\underbrace{\frac{1}{1-a_n}}_{n\text{th place}} ,0, \ldots )$ is more direct.
For any non-negative $n$, let $e (n)$ be the sequence such that $e (n)_i = 0$ if $i \neq n$, and $e (n)_n = 1$. These sequences are eigenvectors for the operator $T$, as:
$T e(n) = (1-a_n) e(n).$
This implies that $1-a_n$ belongs to the spectrum $\sigma(T)$ of $T$ for all $n$. Hence, $\sigma (T)$ has an accumulation point at $1$, which can't happen if $T$ is compact.
Edit: actually, you can show the following stronger properties:
Prove it directly. The bounded sequence $e_n=(\underbrace{0 \ldots 1}_{n\text{th place}} 0 \ldots )$
is mapped to a sequence with no Cauchy subsequences.
The idea which consists of looking at the inverse works, because an invertible linear bounded operator between two infinite dimensional Banach spaces is never compact.