 Solution

Suppose first that $f-u^{\ast }$ is orthogonal to ${\mathcal{U}}_n$. Every $u\in {\mathcal{U}}_n$ can be written $u=u^{\ast }+v$ with $v\in {\mathcal{U}}_n$. The Pythagorean theorem in an inner-product space givesThus $u^{\ast }$ is a best approximation.

Conversely, if $u^{\ast }$ minimizes the distance, then for every $v\in {\mathcal{U}}_n$ the quadratichas its minimum at $t=0$. Differentiating there gives $q^{\prime }(0)=-2(f-u^{\ast },v)=0$ in the real case; varying real and imaginary parts gives the complex case. Therefore