Let $X$ be a Hilbert space and $Y \neq X$ a closed subspace of $X$. How can I prove that, for all $x \in X$
$$\min\left\{||x-y||:y\in Y \right\}=\max\left\{|(x|z)|: z \in Y^{\perp}, ||z||=1\right\}?$$
Of course the expression at first member is the projection of $x$ on $Y$.
Use the fundamental theorem that $X = Y \oplus Y^\perp$. Any element $x \in X$ has the unique form $x = m+n$ where $m \in Y$ and $n \in Y^\perp$. If $n = 0$ then both the min and max are zero, so assume $n \not= 0$.
For any $y \in Y$ the Pythagorean theorem gives you $$ \|x-y\|^2 = \|(m+n) - y\|^2 = \|m-y\|^2 + \|n\|^2 \ge \|n\|^2.$$ Thus $$ \min \left\{ \|x-y\| : y \in Y \right\} = \|n\|$$ attained when $y = m$.
On the other hand if $z \in Y^\perp$ and $\|z\| = 1$ then $$|\langle x, z\rangle| = |\langle m+n,z \rangle| = |\langle n,z \rangle| \le \|n\| \|z\| = \|n\|$$ so that $$ \max\left\{ |\langle x,z \rangle| : z \in Y^\perp,\ \|z\| = 1 \right\} = \|n\|$$ attained when $z = \dfrac n{\|n\|}$.
