 Solution

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation in the finite-dimensional linear subspace, then for every $v$ in it and every real $t$,A nonzero real linear coefficient would make this negative for sufficiently small $t$ of the appropriate sign. Hence $\mathrm{Re}(r,v)=0$. In a complex inner product space, apply the argument to $iv$ too; the imaginary part also vanishes. Thus $(x-u^{\ast },v)=0$ for every $v$ in the subspace.

Conversely, if that orthogonality holds, every other candidate has the form $u^{\ast }+v$, and the Pythagorean theorem in the inner product norm givesEquality requires $v=0$. ThereforeThe next part constructs this orthogonal projection explicitly. No completeness of the whole inner product space is required for projection onto a finite-dimensional subspace.