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?
Best Answer
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 $X$.