To Show Closedness of a Graph in an Application of Closed Graph Theorem

Question

Here's an old exam question I am struggling with:

Let E be a Banach space and $ (x_n)_{n \in N} \subset E $ such that $ \sum_{n=1} ^{\infty} | \langle x_n , x^* \rangle | < \infty $ for all continuous linear functionals $ x^* \in E^* $.

Show that then there exists a constant $ C < \infty $ such that $$ \sum_{n=1} ^{\infty} | \langle x_n , x^* \rangle | \leq C||x^*|| .$$

What I know is that I should show that the graph of a linear map $ T: E^* \rightarrow l^1 $ is closed and then the continuity of $ T $ would follow from the closed graph theorem.

But the problem here is that I don't have any idea where to start showing closedness of the graph.

Answer 1

Well, I am not using the Closed graph theorem really but I am using Banach-Steinhaus theorem, which is a corollary of the Uniform Boundedness Principle. (I think it is possible to deduce each of them using the other.)

As you said we will define a continuous map from $E^* \to \ell ^1$.

Define $T_{k} : E^* \to \ell ^1$ by $T_k(x^*) = (x^*(x_1) , \ldots, x^*(x_k), 0,0 \ldots ) $. It is easy to see that each $T_k$ is linear.

We prove it is bounded. Indeed,

$||T_k(x^*)||_{\ell ^1} = \sum_{i=1}^{k} |x^*(x_i)| \le \sum_{i=1}^{k} ||x^*|| ||x_i|| = \left( \sum_{i=1}^{k} ||x_i|| \right) ||x^*||$.

Now, by your hypothesis, $T_k(x^*)$ converges as $k \to \infty$ for each $x^* \in E^*$.

Therefore, by BST, $T(x^*) := \lim _{k \to \infty} T_k(x^*)$ is a bounded linear operator.

This gives you the conclusion.