How can I find $\max \langle x,q \rangle$ where $x$ is fixed in a Hilbert space $H$ and $q$ runs over $ A^\bot$ with $\|q\|=1$, $A$ a proper nontrivial subspace? (Here we assume $x \notin A, A^\bot$.)
Is $q$ the projection of $x$ onto $A^\bot$ (and then normalized)? How can I prove this? Any essential suggestion?
Additional consideration : we consider both $\mathbb{C}$ and $\mathbb{R}$ for scalar fields.
In real Hilbert, $A^\perp$ is closed subspace so that we have $$ H = A^\perp \oplus W,\ (q\in )W^\perp=A^\perp $$
Hence $x=x_{w} + x_{w^\perp} $ so that $$ <x,q>=<x_{w^\perp},q > $$
Define $$ f : W^\perp \rightarrow {\bf R},\ f(q)=<q,x_{w^\perp}> $$
Then $$| f(q)| \leq \parallel q \parallel \parallel x_{w^\perp} \parallel $$ so that $${\rm max}_{\parallel q \parallel =1} \ |f(q)| \leq \parallel x_{w^\perp} \parallel $$
Let $q= \frac{ x_{w^\perp} }{\parallel x_{w^\perp} \parallel }$ Then $$f(q) = \parallel x_{w^\perp} \parallel $$