For $(H, \langle\cdot,\cdot\rangle)$ a separable real Hilbert space and $K\subset H$ compact, let $(u_k)$ be a sequence in $H$ such that $(\langle u_k, v\rangle)$ converges for each $v\in K$.
Are there [non-trivialising] conditions on $(u_k)$ and $K$ which imply that
$$\tag{1} \exists \ u\in H \quad \text{with} \quad \lim_{k\rightarrow\infty}\langle u_k, \cdot\rangle = \langle u, \cdot\rangle \ \text{ pointwise on } K \ ? $$
What follows is an example that the uniform convergence on $K$ does not suffice. Let $\{e_j\}_{j=1}^\infty$ be an orthonormal basis. Consider $$v_n=\sum_{j=1}^n 2^{-j}e_j,\quad v=\sum_{j=1}^\infty 2^{-j}e_j$$ and $K= \{v_n\,:\,n\in\mathbb{N}\}\cup \{v\} .$ Then $K$ is compact. For $u_k=\sum_{j=1}^ke_j$ we have $$\lim_k\langle u_k,v_n\rangle =\sum_{j=1}^n2^{-j},\quad \lim_k\langle u_k,v\rangle =1\quad (*)$$ However there is no element $u$ such that $$\langle u,v_n\rangle =\sum_{j=1}^n2^{-j},\quad n\in \mathbb{N}\quad (**)$$ Indeed if $u=\sum_{j=1}^\infty a_je_j,$ then $(**)$ implies $a_j=1$ for all $j.$ Moreover the convergence in $(*)$ is uniform as $$0\le \sum_{j=1}^n2^{-j}-\langle u_k,v_n\rangle =\begin{cases} \sum_{j=k+1}^n2^{-j} & k<n\\ 0 & k\ge n \end{cases} \\ \le \sum_{j=k+1}^\infty 2^{-j}=2^{-k}$$
The same holds for the larger compact set $$K=\{v\,:\,|\langle v,e_j\rangle |\le \alpha_j\}$$ for a fixed square summable nonnegative sequence $\alpha_j$ and the same choice of $u_k.$