Suppose $H,K$ are separable Hilbert spaces and $A : H \to K$ is a closed, densely defined, unbounded operator with domain $D(A)$. The following fact is often useful:
Proposition. Suppose $x_n \in D(A)$, $x_n \to x$ in $H$, and $\{A x_n\}$ is bounded in $K$. Then $x \in D(A)$ and $A x_n \rightharpoonup A x$ weakly in $K$.
Proof. Since bounded subsets of $K$ are weakly precompact and metrizable, we may pass to a subsequence and assume that $\{A x_n\}$ converges weakly in $K$ to some $y$. Suppose $k \in D(A^*)$. Then $(x, A^* k)_H = \lim (x_n, A^* k)_H = \lim (A x_n, k)_K = (y,k)_K$. This means that $x \in D(A^{**})$ and $A^{**} x = y$. But since $A$ is closed and densely defined, $A^{**}=A$. Moreover, we have just shown that $(A x_n, k)_K \to (y,k)_K = (Ax, k)_K$ for all $k \in D(A^*)$. Since $A$ is closed, $D(A^*)$ is dense in $K$, and $\{A x_n\}$ is bounded, so it follows from the triangle inequality that $(A x_n, k)_K \to (Ax, k)_K$ for all $k \in K$, i.e. $A x_n \rightharpoonup Ax$ weakly in $K$. Since the same holds if we passed to a different weakly convergent subsequence, the original sequence $\{A x_n\}$ must itself converge weakly to $Ax$.
I have two questions.
For such a simple statement, it seems that this proof uses a lot of machinery. In particular, the weak compactness seems like a big hammer. There are also a lot of properties of adjoints, which, while elementary, take some work to establish. Does anyone know a simpler proof of the proposition?
If I replace $H,K$ by Banach spaces, is the theorem still true? If the spaces are reflexive, it seems probable (but I would have to look up that all the relevant facts used are still true). What if $H,K$ are not reflexive?