# Proof check: the image of the closed unit ball under a compact operator is closed if the domain is reflexive.

The task at hand is the following:

Let $K: X \to Y$ be a compact operator, $X$ and $Y$ Banach spaces, where $X$ is reflexive in addition. Claim: $K(B)$ is a closed subset of $Y$ where $B$ denotes the closed unit ball in $X$.

I'd like someone to go over the proof and point out mistakes, what could have been done better, faster or where I should have been more precise.

Proof:

Let $K(x_n)$ be a sequence in $K(B)$ (i.e. $x_n$ a sequence in $B$) that converges in $Y$. Then by the Eberlein theorem we obtain that every bounded sequence $x_n$ in $X$ has a weak convergent subsequence $x_{n_i}$ with weak limit $x \in X$. Since K is a compact operator we can then extract a subsequence $x_{n_{i_j}}$ such that $K(x_{n_{i_j}})$ converges to some $y \in Y$.

Now we are gonna show that $K(x)=y$:

Let $y^* \in Y^*$:

$$\langle y^*,y\rangle=lim_{j \to \infty}\langle y^*,K(x_{n_{i_j}})\rangle=\langle K^*y^*,x\rangle=\langle y^*,K(x)\rangle$$

where in the first step I used that the pairing $\langle \cdot,\cdot\rangle$ is continuous, in the second I use the property of the dual operator and the weak convergence and finally again the dual operator formula.

Since $$\langle y^*,y-K(x)\rangle = 0 \space \forall y^* \in Y^*$$

we can conclue by Hahn-Banach that $y=K(x)$, hence $K(B)$ is closed. q.e.d.

• Simpler, IMO: $X$ is reflexive, hence $B$ is weakly compact. $K$ is continuous, hence it is weakly continuous, so $K(B)$ is weakly compact. Hence $K(B)$ is weakly closed. $B$ is convex (and balanced), $K$ is linear, hence $K(B)$ is convex (and balanced), so it is also norm-closed. – Daniel Fischer Jan 15 '16 at 17:13
• Your proof is correct, but there's no need for the sub-subsequence $(x_{n_{i_j}})$. You started with a sequence $(x_n)$ such that $K(x_n)$ converges, say to $y_0$. Then $K(x_{n_i}) \to y_0$ for every subsequence $(x_{n_i})$ of $(x_n)$. So after you picked the weakly convergent subsequence by Eberlein, you can work with that - and that shows that the compactness of $K$ is irrelevant. – Daniel Fischer Jan 16 '16 at 13:36
• As for the equivalence of weak continuity and norm continuity of linear operators, I'm not sure how intuitive one would call it. When you're looking at dual pairings and the topologies induced by them, you have a simple characterisation of weak continuity, a linear map $A \colon (E_1,\sigma(E_1,F_1)) \to (E_2,\sigma(E_2,F_2))$ is continuous if and only if $A^t(F_2) \subset F_1$, where $A^t \colon E_2^+ \to E_1^+$ is the transpose of $A$ - $E_i^+$ is the algebraic dual of $E_i$, the space of all linear forms. If the $E_i$ are Hausdorff locally convex spaces, and $F_i$ is the topological – Daniel Fischer Jan 16 '16 at 14:05
• dual of $E_i$, it is clear that for every continuous (with respect to the original topologies) linear $A\colon E_1 \to E_2$ we have $A^t(F_2) \subset F_1$, since the composition of continuous maps is continuous. That gives the "continuous $\implies$ weakly continuous" direction. For the other direction we note that $(A^t)^t(E_1) = A(E_1) \subset E_2$, so $A^t \colon (F_2,\sigma(F_2,E_2)) \to (F_1,\sigma(F_1,E_1))$ is continuous. Thus $A^t$ maps $\sigma(F_2,E_2)$ bounded/compact subsets of $F_2$ to $\sigma(F_1,E_1)$ bounded/compact subsets of $F_1$. Since the Mackey topology $\tau(E_i,F_i)$ – Daniel Fischer Jan 16 '16 at 14:05
• and the strong topology $b(E_i,F_i)$ are determined by $\sigma(F_i,E_i)$-compact resp. $\sigma(F_i,E_i)$-bounded sets, that implies that $(E_1,\sigma(E_1,F_1)) \to (E_2,\sigma(E_2,F_2))$ continuity implies $(E_1,\tau(E_1,F_1)) \to (E_2,\tau(E_2,F_2))$ continuity and $(E_1,b(E_1,F_1)) \to (E_2,b(E_2,F_2))$ continuity. But for a normed space $E$, the norm topology is equal to $\tau(E,E^{\ast})$ and $b(E,E^{\ast})$, so for normed spaces, weak continuity implies norm-continuity. – Daniel Fischer Jan 16 '16 at 14:06

I think you don't need the assertion that $K$ is compact in your proof, but that it is enough for $K$ to be bounded:
You assume that $y := \lim_{n \rightarrow \infty} K(x_n)$ exists for a sequence $(x_n)_{n \in \mathbf{N}} \subset B$. Now if $X$ is reflexive, we may extract a subsequence $(x_{n_i})_{i \in \mathbf{N}}$ which converges weakly to $x \in B$. Then: $$\langle y^*, y \rangle = \lim_{i \rightarrow \infty} \langle y^*, K(x_{n_i}) \rangle = \lim_{i \rightarrow \infty} \langle K^* y^*, x_{n_i} \rangle = \langle y^*, K(x) \rangle$$ since $K^* y^* \in X^*$. Hence $y = K(x) \in K(B)$.
I was thinking, whether compactness of $K$ could help you to prove the converse, but there's the following counterexample: Consider the non-reflexive space $\ell^1$ and the operator $K: \ell^1 \rightarrow \mathbf{R}$, $(x_n)_{n \in \mathbf{N}} \mapsto x_1$, which is compact, for it is of finite-rank; the image of the unit ball is $K(B) = [-1, 1]$.
• Didn't you mean 'by continuity of $K$'? I like the your example even if it appears strange to me that one does not need the compactness of $K$ – noctusraid Jan 16 '16 at 8:08