continuous function to continue linear functional in Hilbert space

Question

Let $H$ be a Hilbert space infinite dimensional over the complex numbers $\mathbb C$.

Let $S$ be a compact subset of $H$.

Let $V=\overline{\operatorname{span}(S)}$ be the closure of linear span of $S$.

Let $g:S \to \mathbb C$ be a continuous function or rather for each sequence $s_n$ of $S$ such that $s_n \to s_0$ ($H$ metric) then $g(s_n) \to g(s_0)$.

My question is:

Is it always possible to extend the $g$ function to a 'continue' linear functional on $V$?

Thanks.

Answer 1

Let $H=\ell^2(\mathbb N)$, $S=\{e_1,e_1+\frac1{n^2}e_n\mid n\in\mathbb N,n>1\}$, $g(e_1+\frac1{n^2}e_n)=1+1/n$, $g(e_1)=1$.

You have:

1. $S$ is compact, since any sequence that has an infinite amount of distinct elements must have a subsequence converging to $e_1$.
2. $S$ is linearly independent ($e_i$ is not a linear combination of any of the other $e_j$).
3. $g$ is continuous, since the only accumulation point is $e_1$ so if $e_1+\frac{1}{n_k^2}e_{n_k}\to e_1$ you must have $n_k\to\infty$ and $g(e_1+\frac1{n_k^2}e_{n_k})=1+\frac1{n_k}\to1=g(e_1)$.

However any linear extension $g'$ must have: $$g'(e_n)=g'\left(n^2(e_1+\frac{1}{n^2}e_n-e_1)\right)=n^2\left(g'(e_1+\frac1{n^2}e_n)-g'(e_1)\right)=n^2+n-n^2=n$$ In other words $\sup_{x\in H,\|x\|≤1}|g'(x)|$ is infinite. This is equivalent to $g'$ not being continuous.