Prove that $v \in U^\perp $. 
Let $U$ be a subspace of $V$. Assume that there exists $v\in V$ such that $\langle v,u\rangle +\langle u,v\rangle \le \langle u,u\rangle \forall u\in U$
Prove that $v \in  U^\perp $.

Let $u\in U$ then we need to show that $\langle u,v\rangle =0$
Now $\langle v,u\rangle +\langle u,v\rangle \le \langle u,u\rangle\implies 2\Re \langle u,v\rangle\le \langle u,u\rangle$
How to show from here that $\langle u,v\rangle =0$?
Please help
 A: We take the Chessanator hint, and note that $U\cap U^{\perp}=\{0\},$ the zero vector. If $v=0,$ then we are done trivially because $0\in U^{\perp}.$ But now, suppose, by way of establishing a contradiction, that $v\in V,$ and $v\not=0$. Then, by assumption, we can write $v=u+u^{\perp},$ where $u\in U$ and $u^{\perp}\in U^{\perp}$. It must be that
$$\langle v,u\rangle+\langle u,v\rangle=\langle u+u^{\perp},u\rangle+\langle u,u+u^{\perp}\rangle=\langle u,u\rangle+\langle u,u\rangle =2\langle u,u\rangle.$$
If $u=0,$ we are done and $v\in U^{\perp}.$ If $u\not=0,$ then we have a case where $\langle v,u\rangle+\langle u,v\rangle > \langle u,u\rangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
A: Suppose there is some $u\in U$ such that $\langle u,v\rangle\neq 0$. Pick $\lambda\in\mathbb{C}$ and take $w=\lambda u$, such that $\langle w,v\rangle>0$ and $\|w\|=1$.
Now let $r>0$ and consider
$$\langle v,rw\rangle+\langle rw,v\rangle\leq \|rw\|^{2}=r^{2}\Leftrightarrow r^{2}-2r\langle v,w\rangle\geq0$$
which only holds for $r\leq0$ and $r\geq 2\langle v,w\rangle$. So for $r=\langle v,w\rangle$ we find
$$\langle v,rw\rangle+\langle rw,v\rangle=2\langle v,w\rangle^{2}\geq \langle v,w\rangle^{2}=\|rw\|^{2}$$
which is a contradiction.
