Suppose you have a complex inner product space $V$, a subspace $S$, and some vector $v\in V$ such that $$\langle v,s\rangle+\langle s,v\rangle\leq\langle s,s\rangle$$ for all $s\in S$. How can you determine $v\in S^\perp$?
I jotted down that for a given $\epsilon>0$ and $s\in S$, there exists $c\in\mathbb{C}$ depending on such $\epsilon$ and $s$ such that $\langle cs,cs\rangle=|c|^2\|s\|<\epsilon$. Then $$ \begin{align*} \langle v,cs\rangle+\langle cs,v\rangle &=\bar{c}\langle v,s\rangle+c\langle s,v\rangle\\ &=\bar{c}\langle v,s\rangle+c\overline{\langle v,s\rangle}=2\Re(\bar{c}\langle v,s\rangle)\\ &\leq \langle cs,cs\rangle=|c|^2\|s\|<\epsilon \end{align*} $$
My hope was to somehow show $\Re\langle v,s\rangle$ and $\Im\langle v,s\rangle$ equal $0$, but it seems to be going in the wrong direction. Is there a more useful approach?
Hint: Let $s\in S$ be a nonzero vector. Then $\varepsilon \alpha s\in S$ for every $\varepsilon>0$ and every $\alpha\in\mathbb{C}$. Hence $\langle v,\varepsilon \alpha s\rangle+\langle \varepsilon \alpha s,v\rangle\leq \langle \varepsilon \alpha s,\varepsilon \alpha s\rangle$ for all $\varepsilon>0$. Argue that $\mathrm{Re}(\langle \alpha s,v\rangle)=0$. As $\alpha$ is arbitrary, argue further that $\langle s,v\rangle=0$.