Help required with question about closed unit ball in Hilberts space and proving the projection formula

Question

I've this question that I intend to prove and any help will be appreciated

Answer 1

In $\mathbb{H}$ the projection of $x$ onto the closed unit ball $B_1[0]$ is $w\in B_1[0]$ such that $\|x-w\|=\min_{v \in B_1[0]} \|x-v\|$. Evidently if $x\in B_1[0]$ then $w=x$. So what you have to prove is that $w=\frac{x}{\|x\|}$ for $x\notin B_1[0]$.

We have to prove that $\|x-v\|\ge\|x-w\|\ \forall v\in B_1[0]$ where $w=\frac{x}{\|x\|}$. Lets start: $$\|x-v\|\ge\|x-w\| \iff \|x-v\|^2\ge\|x-w\|^2$$ $$\iff \langle x-v,x-v\rangle\ge\langle x-w,x-w\rangle$$ $$\iff \langle x,x\rangle-2\langle x,v\rangle+\langle v,v\rangle\ge\langle x,x\rangle-2\langle x,w\rangle+\langle w,w\rangle$$ $$\iff -2\langle x,v\rangle+1\ge-2\langle x,w\rangle+1 \iff \langle x,w\rangle\ge\langle x,v\rangle$$ $$\iff \langle x,\frac{x}{\|x\|}\rangle\ge\langle x,v\rangle \iff \frac{1}{\|x\|}\langle x,x\rangle\ge\langle x,v\rangle$$ $$\iff \|x\|\ge\langle x,v\rangle$$ But the last inequality is a result of Cauchy's $\|x\|\|v\|\ge\langle x,v\rangle$ where $\|v\|=1\square$.