Prove that $\sup\limits_{\alpha \in D} \|x_{\alpha}\| \lt \infty.$

Question

Let $X$ be a complex normed linear space. Let $D$ be a directed set and $\{x_{\alpha} \}_{\alpha \in D}$ be a net in $X$ converging to $x \in X$ with respect to the weak topology i.e. for all $\varphi \in X^*$ we have $\varphi (x_{\alpha}) \to \varphi (x).$ If $\sup\limits_{\alpha \in D} \left \lvert \varphi (x_{\alpha} ) \right \rvert \lt \infty,$ for any $\varphi \in X^*$ then prove that $\sup\limits_{\alpha \in D} \|x_{\alpha} \| \lt \infty.$

Attempt $:$ Consider the collection $\{J_{\alpha} \}_{\alpha \in D}$ in $X^{**}$ defined by $J_{\alpha} (\varphi) = \varphi (x_{\alpha}),$ $\varphi \in X^*.$ Then it is easy to see that $\|J_{\alpha}\| \leq \|x_{\alpha}\|.$ Also By Hahn-Banach theorem there exists $\varphi_{\alpha} \in X^*$ for each $\alpha \in D$ with $\|\varphi_{\alpha}\| = 1$ such that $\varphi_{\alpha} (x_{\alpha}) = \|x_{\alpha}\|.$ This shows that $\|J_{\alpha}\| \geq |J_{\alpha} (\varphi_{\alpha})| = \|x_{\alpha}\|,$ proving that $\|J_{\alpha} \| = \|x_{\alpha} \|.$ Now by the given hypothesis it follows that for any $\varphi \in X^*$ we have $\sup\limits_{\alpha \in D} J_{\alpha} (\varphi) \lt \infty.$ Hence by uniform boundedness principle it follows that $$\sup\limits_{\alpha \in D} \|x_{\alpha}\| = \sup\limits_{\alpha \in D} \|J_{\alpha}\| \lt \infty$$ as required.

In the above proof I didn't use anywhere the weak convergence of the net $\{x_{\alpha} \}_{\alpha \in D}$ and also didn't see any special role of the directed set $D$ as well. Are they of no importance here? Any suggestion regarding this will be highly appreciated.

Thanks in advance.

Answer 1

Without loss of generality, we may assume that $X$ is a Banach space. For, let $\bar{X}$ be the completion of $X$. Recall that every bounded linear function $\varphi\in X^{\ast}$ has a unique continuous extension to $\bar{\varphi}\in\bar{X}^{\ast}$. On the other hand, given $\varphi\in\bar{X}^{\ast}$, we have that $\overline{\varphi\vert_{X}}=\varphi$, i.e., the continuous extension of the restriction of $\varphi$ on $X$ is itself. Hence, there is a one-one correspondance between $X^{\ast}$ and $\bar{X}^{\ast}$. Now, it is clear that for each $\varphi\in\bar{X}^{\ast}$, we have that $\sup_{\alpha\in D}|\varphi(x_{\alpha})|<\infty.$

Assume that $X$ is complete. Consider the canonical embedding $X\rightarrow X^{\ast\ast},$ $x\mapsto\hat{x}$, where $\hat{x}(\varphi)=\varphi(x)$, $\varphi\in X^{\ast}$. Recall that the canonical embedding is an isometric homomorphism, and hence $||x||=||\hat{x}||$. The given condition implies that: $\{\hat{x}_{\alpha}\mid\alpha\in D\}$ is a family of bounded linear functionals and for each $\varphi\in X^{\ast}$, $\sup_{\alpha\in D}|\hat{x}_{\alpha}(\varphi)|<\infty$. By Uniform Boundedness Principle, we have $\sup_{\alpha\in D}||\hat{x}_{\alpha}||<\infty$. Note that weak convergence is irrelevant.