Proving that $\|z\| \leq \|z-y\|$ for all $y\in F\implies z \perp F$

Question

I'm having some trouble with the following exercise:

Let $F$ be a subspace of a complex inner product space $E$, and $z\in E$. Prove that the following are equivalent:

1. $\|z\|\leq\|z-y\|$ for all $y\in F$
2. $z \perp F$
3. $d(z,F) = \|z\|$

I was able to prove that $(2) \implies(3)$, $(3)\implies(1)$ but now I'm having a lot of trouble proving that $(1)\implies(2)$. I remember proving this in linear algebra, but the last time I proved this, I used the fact that $F$ was a Real inner product space with finite dimensions, and this is no longer the case. How can this be

Answer 1

Let $y\in F$ and $a\in \mathbb{C}$ with $|a|=1$ (we shall choose the exact value of $a$ in a bit). For every $t\in \mathbb{R}$, $(1)$ gives $$\|z\|^2 \leq \|z-tay\|^2 = \|z\|^2-ta\langle z, y\rangle - t\overline{a}\langle y, z\rangle + t^2\|y\|^2$$

If we choose $a=\frac{\langle y, z\rangle}{|\langle y, z\rangle|}$ and simplify we get that for every $t\in \mathbb{R}$ it holds $$\|y\|^2 t^2 - 2t|\langle z, y\rangle| \geq 0$$

For this to happen it is necessary that $\langle z, y\rangle = 0$. Since $y$ was arbitrary, we have $z\perp F$.