Let $(H, \langle\cdot,\cdot\rangle)$ be a separable real Hilbert space, and $\mathcal{S}$ be an infinite subset of $H$ such that the elements of any finite subset of $\mathcal{S}$ are linearly independent. Any map $f : \mathcal{S}\rightarrow\mathbb{R}$ then admits a unique linear extension $\bar{f} : V\rightarrow \mathbb{R}$ to the subspace $V:=\mathrm{span}(\mathcal{S})$ of $H$.
Provided that $\max_{s\in\mathcal{S}}|f(s)| \leq C$ for some $C>0$, under which conditions on $f$ and $\mathcal{S}$ can we conclude that the linear extension $\bar{f}$ is a bounded functional on $V$?
(It is clear that $\bar{f}$ will not be bounded in general, as the example $(H, \mathcal{S}, f):=(\ell_2(\mathbb{N}), \{e_i\mid i\in\mathbb{N}\}, \langle (1,1,1,\ldots), \cdot\,\rangle_{\ell_2})$ shows.)
Well, an easy answer comes from Riesz representation theorem: if you can write $f$ as $f(x)=\langle y,x\rangle$ for some $y\in H$, then the extension is clearly bounded on $V$.
Also, if $\overline f$ is bounded on $V$, by Hahn-Banach theorem, it can be extended to a bounded map $g:H\to \mathbb R$, which then by Riesz theorem has to be represented by testing against an element $y$ of $H$ (and also, by applying the same argument to $\overline V$, which is a Hilbert space, there exists a unique of such $y$ that belongs to $\overline V$, where uniqueness follows from the density of $V$ in $\overline V$ and the bijective correspondence given by Riesz repr. theorem).
So, the above condition is both necessary and sufficient: $f$ extends to a bounded function on $V$ if and only if $f(x)=\langle y,x\rangle$ for some $y\in H$ (if and only if $f(x)=\langle y,x\rangle$ for some $y\in \overline V$).