Functional Analysis, Sequence of continuous linear functionals

Question

Let {${\phi_n}$}$_{n=1}$ be sequence of continuous linear functionals $ X \to K $ on a Banach Space $X$ such that {$\phi_n(x)$}$_{n=1}$ is convergent for every $ x \in X$ . Show that the sequence {$\phi_n(x_n)$}$_{n=1}$ is convergent whenever {$x_n$}$_{n=1}$ is convergent in $X$.

Answer 1

I assume $K$ is either $\Bbb C$ or $\Bbb R$. Let $x\in X$ be the limit of $x_{n}$ and let $k\in K$ be the limit of $\phi_{n}(x)$. We want to prove that $k$ is the limit of $\phi_{n}(x_{n})$.

We have:

\begin{align*} &\vert\phi_{n}(x_{n})-k\vert \\ &= \vert\phi_{n}(x_{n})-\phi_{n}(x)+\phi_{n}(x)-k\vert\\ &\le \vert \phi_{n}(x_{n})-\phi_{n}(x)\vert+\vert \phi_{n}(x)-k\vert \tag{triangle inequality in $K$}\\ &= \vert\phi_{n}(x_{n}-x)\vert+\vert\phi_{n}(x)-k\vert \tag{as $\phi_{n}$ are linear functionals}\\ &\le \Vert\phi_{n}\Vert\cdot\Vert x_{n}-x\Vert_{X}+\vert \phi_{n}(x)-k\vert \tag{as $\phi_{n}$ are linear maps on $X$} \end{align*}

Note that for any $n$, $\Vert\phi_{n}\Vert<\infty$ since the $\phi_{n}$ are continuous. Thus, as $x_{n}\to x$ and $\phi_{n}(x)\to k$ by construction, we only need to prove that there exists a bound $M$ independent on $n$ such that $\Vert\phi_{n}\Vert\le M$.

But this is a consequence of the uniform boundedness principle, which we can apply because we have the appropriate hypothesis.

Indeed, the fact that $\phi_{n}(x)$ converges to some $k_{x}$ for any $x\in X$ directly implies that $\sup_{n}\vert\phi_{n}(x)\vert<\infty$: fix $\epsilon>0$. There exists $N_{\epsilon}$ such that for any $n\geq N_{\epsilon}$, we have $\vert\phi_{n}(x)\vert\le \epsilon+\vert k_{x}\vert$, and thus $\sup_{n}\vert\phi_{n}(x)\vert\le\max\{\vert\phi_{N}(x)\vert,\epsilon+\vert k_{x}\vert\,:\, N< N_{\epsilon}\}$.