Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

up vote 2 down vote accepted

You could argue as follows:

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.

This proof works in any reflexive Banach space, and doesn't use adjoints. But you need the result that weak and norm closures of convex sets coincide (I always forget who to attribute this result to-- it follows readily from the geometric form of the Hahn-Banach Theorem).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.