Exercise :

Let $X,Y$ be Banach spaces with $\dim X = + \infty$ and $A \in \mathcal{L}(X,Y)$ such that : $$\|A(x)\|_Y \geq c\|x\|_X \; \; \forall x \in X \; \text{and} \; c>0$$ Can the operator $A$ be compact ?

Attempt :

First of all, we know that an operator is compact if it transfers bounded sets to relatively compact sets (which means that they have a compact closure).

Let's assume now that $A$ is compact. Since $X$ is infinite dimensional, the unit ball $B_1^X$ is bounded (but not totally bounded). That would mean that $\overline{A(B_1^X)}$ should be compact. But, from the inequality relation gives, it would be :

$$\|A(B_1^X)\|_Y \geq c\|B_1^X\|_X \implies \|A(B_1^X)\|_Y \geq c' >0$$

But $c'$ could be as large as we'd like and thus the quantity $\|A(B_1^X)\|_Y$ is not bounded, which also means that the $\|\overline{A(B_1^X)}\|_Y$ is also not bounded, thus \overline{A(B_1^X)} is not compact (???).

Question : Is my intuition and especially my final argument correct ? If not, what other way could I approach that specific problem ?

  • $\begingroup$ Are you sure it says for all $c>0$, because such operator is unlikely to exist, because every $x$ has the property $\|Ax\|=\infty$... Maybe it should say there exists some $c>0$ for which this and that holds. $\endgroup$ – Shashi Feb 23 at 11:06

You can do this by definition. Since $X$ is infinite-dimensional, there exists a sequence $\{x_n\}$ in the unit ball $B_X$ of $X$ that has no convergent subsequence; thus there exists $\delta>0$ such that $\|x_k-x_j\|\geq\delta$ if $k\ne j$. Then $$ \|Ax_k-Ax_j\|=\|A(x_k-x_j)\|\geq\,c\,\|x_k-x_j\|\geq\delta c. $$ So the sequence $\{Ax_j\}$ admits no convergent subsquence, and thus $A(B_X)$ is not precompact, and $A$ is not compact.

  • $\begingroup$ Hi, thanks a lot for your input. Shouldn't it be $c$ instead of $1/c$ by the way ? $\endgroup$ – Rebellos Feb 23 at 18:51
  • $\begingroup$ Indeed; edited. $\endgroup$ – Martin Argerami Feb 23 at 19:12
  • 1
    $\begingroup$ I moved the accepted mark because I think this answer is more fundamentaly suited to the knowledge I have and the style of lesson we follow. Essentialy, this is kinda like what I tried to work around, but got mixed up! Thanks for the perfect elaboration ! $\endgroup$ – Rebellos Feb 23 at 19:14

Your attempt does not make too much sense to me. What is the norm of a subset even supposed to mean? If $A$ is bounded, then $A(B_1^Y)$ is certainly bounded.

But you are right that $A$ cannot be compact. Let $Z=A(X)$. By assumption $A$ is bijective from $X$ to $Z$. Let $B$ be its inverse. Then $\|B(Ax)\|_X=\|x\|_X\leq \frac 1 c\|Ax\|_Y$, that is, $B$ is bounded. If $A$ were compact, then $\mathrm{id_X}=BA$ would also be compact, which contradicts the fact that $X$ is infinite-dimensional.

  • $\begingroup$ $Z$ is the range of $A$, so $A$ is certainly also surjective from $X$ to $Z$, if that's what you mean. $\endgroup$ – MaoWao Feb 23 at 11:11

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