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

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 ?

• 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. – 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.
• Hi, thanks a lot for your input. Shouldn't it be $c$ instead of $1/c$ by the way ? – Rebellos Feb 23 at 18:51
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.
• $Z$ is the range of $A$, so $A$ is certainly also surjective from $X$ to $Z$, if that's what you mean. – MaoWao Feb 23 at 11:11