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Let $(a_n)$ be a sequence that tends to zero and $0<a_n<1$. Why is then the mapping $$ \ell_2 \rightarrow \ell_2,\ (x_1,x_2,\ldots)\mapsto ((1-a_1)x_1,(1-a_2)x_2,\ldots)$$ not compact ?

Someone said I should look at its inverse, but that didn't helpt me either...

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4 Answers 4

up vote 4 down vote accepted

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.

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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.
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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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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.

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