 Solution

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation, then for every $v\in {\mathcal{U}}_n$ and real $t$, the competitor $u^{\ast }+tv$ satisfiesThe linear coefficient must be zero, since otherwise choosing sufficiently small $t$ of the appropriate sign makes this expression negative. Hence $\mathrm{Re}(r,v)=0$. In a real inner product space this is the desired orthogonality; in a complex inner product space, apply the same argument to $iv$ to show that the imaginary part also vanishes. Thus $(r,v)=0$ for all $v\in {\mathcal{U}}_n$.

Conversely, suppose this orthogonality holds. Every competitor $u\in {\mathcal{U}}_n$ has $u-u^{\ast }\in {\mathcal{U}}_n$. The Pythagorean identity then givesEquality forces $u=u^{\ast }$. We have proved the characterization, as well as uniqueness:This is the orthogonal projection onto a finite-dimensional subspace. Completeness of the ambient space is not needed. Existence follows, for example, by solving the invertible Gram matrix system in a basis of the finite-dimensional linear subspace.