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...
Why is this operator not compact?
4 Answers
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:
- $I-T$ is compact (which also implies that $T$ is non-compact);
- $\sigma (T) = \{1-a_n: \ n \in \mathbb{N}^*\} \cup \{1\}$.
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0How does $I - T$ being compact imply that $T$ is non-compact? – 2016-11-28
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0@csss: the space is infinite-dimensional. Then $I-T$ being compact imply that $1$ is a not a pole of finite multiplicity for the resolvent of $T$, whence $T$ is not compact. – 2016-12-09
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.
- We check that the inverse operator $T^{-1}$ is defined as $$T^{-1}((x_n)_{n\geq 1})=\left(\frac 1{1-a_n}x_n\right)_{n\geq 1},$$ which is bounded from $\ell^2$ to $\ell^2$.
- If $S_1$ is a compact operator and $S_2$ is bounded then $S_1S_2$ and $S_2S_1$ are compact.
- If $T$ were compact, then the identity operator would be compact. But it's not possible because the unit ball is not strongly compact.