= Solution
Suppose first that $f-u^*$ is <orthogonal> to $\mathcal U_n$. Every $u\in\mathcal U_n$ can be written $u=u^*+v$ with $v\in\mathcal U_n$. The <Pythagorean theorem in an inner-product space> gives
$$
\|f-u\|^2
=\|(f-u^*)-v\|^2
=\|f-u^*\|^2+\|v\|^2
\geq\|f-u^*\|^2.
$$
Thus $u^*$ is a best approximation.
Conversely, if $u^*$ minimizes the distance, then for every $v\in\mathcal U_n$ the quadratic
$$
q(t)=\|f-u^*-tv\|^2
$$
has its minimum at $t=0$. Differentiating there gives $q'(0)=-2(f-u^*,v)=0$ in the real case; varying real and imaginary parts gives the complex case. Therefore
$$
\boxed{u^*\text{ is best}
\iff(f-u^*,v)=0\quad\forall v\in\mathcal U_n}.
$$
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