# Domain of closed operator and weak convergence: elementary proof and extensions?

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.

1. 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?

2. 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?

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One the one hand, the statement is simple, however, having an unbounded operator mapping a convergent sequence to something (weakly) convergent is pretty cool. Well, you put in boundedness and obtain weak convergence and use weak compactness - that sounds appropriate.-Whether the statement hold in Banach spaces - I have no idea... – Dirk Nov 23 '11 at 20:40

Again, move to a subsequence and suppose that $Ax_n \rightarrow y$ weakly. For each $n$ let $X_n$ be the convex hull of $\{ x_m : m>n \}$ and let $Y_n$ be the convex hull of $\{ Ax_m : m>n \}$; thus $Y_n = A(X_n)$. So the weak closure of $Y_n$ contains $y$. But $Y_n$ is convex, so the weak and norm closures coincide. Thus we can find $y_n \in X_n$ with $\| Ay_n - y \| < 1/n$.
Clearly $y_n \rightarrow x$ (as everything in $X_n$ will be close to $x$, for $n$ large enough) and now $Ay_n \rightarrow y$ in norm. So as $A$ is closed, $x\in D(A)$ with $Ax=y$.
Again, $x$ doesn't vary with which subsequence we moved to initially, and so actually $Ax_n\rightarrow y$ weakly.