Image of closed unit ball under a compact operator

Question

Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:

$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.

I have proven this when $X$ is a reflexive space.

Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.

In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.

Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.

On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.

But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.

It's easily proved that if the image is compact, the operator is also compact.

But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?

Answer 1

It is false in general that the image of the closed unit ball under a compact operator is closed (and hence compact). Here is an easy example:

Consider $X = C[0,1]$ with the uniform norm, and the compact operator $A \in B(X)$ defined by the formula:

$\displaystyle\qquad Af(x) = \int_0^x f(t)\,dt$.

Compactness of $A$ is easily proven using Arzelà–Ascoli. Our operator $A$ produces an anti-derivative of any input given to it, and the image of the closed unit ball of $X$ under $A$ is the set

$\displaystyle\qquad \{f \in C^1[0,1] \mathrel: f(0)=0,\ \lVert f'\rVert \leq 1\}$

which certainly is not a closed subset of $X$.

Answer 2

This is much easier. Use the definition of compact operator. An operator is compact if and only if the image of a bounded set is precompact. The unit ball is bounded and any other bounded set is inside some $\lambda$ times the unit ball.