If $\mathrm{min}_{u\in U}||v-u||=||v-u_0||$, then $v-u_0\in U^{\perp}$

Question

Problem

Let $U$ be a subspace of an inner product space V (here V may not be finite dimensional). Fix $v\in V$. If there exists $u_0 \in U$ such that $\mathrm{min}_{u\in U}||v-u||=||v-u_0||$, then $v-u_0\in U^{\perp}$

Attempt

If $t\in\mathbb{R}$ and $u\in U$, then we see that the real function \begin{align*} f(t)&=||v-u_0+ut||^2\\ &= ||v-u_0||^2+2t\mathrm{Re}(\langle v-u_0,u\rangle)+t^2||u||^2 \end{align*} has a minimum at $t=0$. Thus $f'(0)=2\mathrm{Re}(\langle v-u_0,u\rangle)=0$. So $\mathrm{Re}(\langle v-u_0,u \rangle)=0$.

Question

I am stuck trying to show that $\mathrm{Im}(\langle v-u_0,u \rangle)=0$. I would be able to do this if I could show that $f(t)=\langle v-u_0+ut,v-u_0-ut \rangle$ has minimum at $t=0$.

Answer 1

I will use A.G's comment.

First, $\mathrm{Re}(-iz)=\mathrm{Im}(z)$. This can be seen if we write $z=a+bi$: \begin{align*} \mathrm{Re}(-i(a+bi)&=\mathrm{Re}(b-ai) \\ &=b\\ &=\mathrm{Im}(z). \end{align*}

The case for real $t$ was proven in my attempt above. Now consider purely imaginary $t$, i.e $t=xi$, where $x\in \mathbb{R}$. Then the same calculation can be done: \begin{align*} \langle v-u_0+uxi, v-u_0+uxi \rangle &= ||v-u_0||^2+2t\mathrm{Re}(-i\langle v-u_0,u\rangle)+t^2||u||^2 \\ &= ||v-u_0||^2+2t\mathrm{Im}(\langle v-u_0,u\rangle)+t^2||u||^2. \end{align*} Again, this function's derivative must vanish at $t=0$ so we find that $\mathrm{Im}(\langle v-u_0,u \rangle)=0$ for all $u \in U$.