Compact operators: why is the image of the unit ball only assumed to be relatively compact?

Question

Recall the definition of compact operators between Hilbert spaces:

An operator $A$ is called compact if the image $A(\mathcal U_H)$ of the unit ball is relatively compact (i.e. its closure is compact) in the norm topology.

However, I seem to be able to "prove" that the image is, in fact, compact, not just relatively compact. What went wrong?

$\square$: We show that $A(\mathcal U_H)$ is sequentially compact. Since the topology is given by a norm / metric, this implies that it is also compact.

Let $y_n = Ax_n$ be a sequence in $A(\mathcal U_H)$. The unit ball in a Hilbert space is weakly compact, so there is a weakly converging subsequence $x_{n_j} \to^w x$ which gives $Ax_{n_j} \to^w Ax$.

On the other hand, by the sequential compactness of the closure of $A(\mathcal U_H)$, we know that a subsequence of $z_k = x_{n_{j_k}}$ has the property $Az_k \to y$ for some $y$ in the Hilbert space, though we do not yet know that $y$ is also in the image of the unit ball.

But we have $z_k \to^w x$ as well, and hence $A z_k \to^w A x$. Since the weak limit must coincide with the norm limit, we have $y = Ax$, and $y \in A(\mathcal U_H)$ already. In other words, the original sequence $y_n$ has a subsequence that converges in the image of the unit ball. $\square$

I feel very stupid for not finding my mistake that must surely be very elementary. What went wrong? A counterexample would probably help my understanding as well.

Answer 1

There is nothing wrong with your argument here, assuming that $\cal U_{\cal H}$ is the closed unit ball. (I imagine that "relative" is included in the definition of compactness since it's needed in the more general setting when defining compact operators between Banach spaces, or when defining compactness of $T$ as "$T(M)$ is relatively compact for any bounded set $M$".)

More generally, if $X$ is reflexive, $M$ is a closed, bounded, and convex subset of $X$ , and $Y$ is a Banach space, then for a compact operator $T:X\rightarrow Y$, the set $T(M)$ is norm-compact in $Y$.

Answer 2

The open unit ball in a Hilbert space is not compact, nor weakly compact. The closed unit ball is weakly compact. Your $x$ might have $||x||=1$. Take any sequence $x_n \in \mathcal{U_H}$ that converges to some $x \in \mathcal{H}$ with $||x||=1$. Then, any subsequence must converge strongly and weakly to $x$ and, running your argument, you get that $y = Ax$ with $x \notin \mathcal{U_H}$, which doesn't tell you that $y$ is in the image of the open unit ball.