Let $X$, $Y$ be normed spaces and $A:X\to Y$ be a linear and continuous operator, which has the property \begin{equation} \exists C>0 \text{ s.t. } \forall x\in X \quad \Vert Ax\Vert\geq C \Vert x \Vert \end{equation} Can A be a compact operator?

I guess, yes it can. But how to show it? I'm confused because $A$ is not bounded, and intuitively I would to use Arzela-Ascoli.

  • $\begingroup$ If $A$ is not bounded it cannot be compact. $\endgroup$ – user42761 Jul 22 '16 at 11:58
  • $\begingroup$ even if $dim(X)<\infty$? $\endgroup$ – MorganeMaPh Jul 22 '16 at 12:00
  • $\begingroup$ I see that you write $A$ is continuous by which most people understand a bounded operator. So what do you exactly mean by bounded ? $\endgroup$ – user42761 Jul 22 '16 at 12:02

In fact, $A$ can't be a compact operator (if $X$ is infinite dimensional). This condition is sometimes referred to as "$A$ is bounded away from $0$".

To show that it can't be compact, it suffices to show that there is a bounded sequence whose image has no convergent subsequence. Note (by the usual application of Riesz's lemma) that there exists a bounded sequence in $X$ such that the distance between any two elements in the sequence has a lower bound.

On the other hand, if $X$ is finite dimensional, then every operator on $X$ is compact. If $X$ is infinite dimensional and $Y$ isn't, then the operator can't have this property.


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