I am trying to prove that:
Let $K$ be a closed konvex subset of a Hilbert space $H$. Let $x_o \in H$, then for all $x \in K$ the following are equivalent:
- $\lVert x_o -x \rVert = inf_{y \in K} \lVert x_0-y\rVert$
- $Re \langle x_0-x,y-x\rangle \leq 0 $ $\forall y \in K$
$1 \Rightarrow 2$: For $y \in K$, we have that $y_t=(1-t)x+ty \in K$. From assumption 1) I get that $\lVert x_0-x\rVert^2 \leq \lVert x_0-y_t \rVert=\langle x_0-x+t(x-y),x_0-x+t(x-y)\rangle=\lVert x_0-x \rVert^2+2 Re\langle x_0-x,t(x-y) \rangle + t^2 \lVert x-y \rVert^2$
This means that I get: $\lVert x_0-x\rVert^2 \leq\lVert x_0-x \rVert^2+2 Re\langle x_0-x,t(x-y) \rangle + t^2 \lVert x-y \rVert^2$ this means we get $0 \leq 2 Re\langle x_0-x,t(x-y) \rangle + t^2 \lVert x-y \rVert^2=t(2 Re \langle x_0-x,x-y \rangle+t\lVert x-y\rVert^2)$
If we assume that $t \neq 0$ we further get:
$2 Re \langle x_0-x,x-y \rangle \leq +t\lVert x-y\rVert^2$
That's how far I got in the proof. Not knowing how to continue, I tried to take a look at how the book I am reading did the prove. But the book did nearly the same calculations as I did and at the point $2 Re \langle x_0-x,x-y \rangle \leq +t\lVert x-y\rVert^2$ notes that Nr2. just follows. My question is, I did assume that $t \neq 0$, and thus I don't see how to conlclude Nr.2 from $2 Re \langle x_0-x,x-y \rangle \leq +t\lVert x-y\rVert^2$, does anybody know why?