If $\|A(x)\|_Y \geq c \|x\|_X$ and $\dim X = + \infty$, can $A \in \mathcal{L}(X,Y)$ be a compact operator?

Question

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 ?

Answer 1

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.

Answer 2

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.