Interior solution of a Variational Inequality in Hilbert Space I'm trying to understand the proof of the following claim:

Let $K \subset \mathbb{R}^n$ be compact and convex and let $$F:K \rightarrow (\mathbb{R}^n)'$$ be continuous. Then there exists an $x\in K$ s.t. the dual pairing $$\langle F(x), y-x \rangle \geq 0 \quad \forall y \in K$$ Further, if $x$ satisfying the above inequality is on the interior of $K$ then $F(x) = 0$.

I understood the existence, but not the proof for $F(x) = 0$ which proceeds as follows:

If $x$ is interior to $K$ then the points $y-x$ describe a neighborhood of the the origin -- ie for any $q \in \mathbb{R}^n$ there exists an $\epsilon \geq 0$ and $y \in K$ s.t. $q = \epsilon(y-x)$ so that:
  $$\langle F(x), q \rangle = \epsilon \langle F(x), y-x \rangle \geq 0 \quad \forall q \in \mathbb{R}^n$$
This implies $F(x) = 0$

I, in particular, don't quite follow the statements in bold.


*

*I don't see any translation to the origin or anything, so this confused me, are they assuming $x$ is the origin or something?

*How does $F(x) = 0$ follow from the inequality.



  
*
  
*I don't see any translation to the origin or anything, so this confused me, are they assuming $x$ is the origin or something?
  

They chose $x$ as the origin of a new coordinate system by the change of variables $y\mapsto y-x$ i.e. they simply translate the starting coordinate system.


  
*How does $F(x)=0$ follow from the inequality?
  

Since $F\in(\Bbb R^n)^\prime$, the duality pairing $\langle\:,\,\rangle:(\Bbb R^n)^\prime\times \Bbb R^n\to\Bbb R$ is simply the euclidean scalar product in $\Bbb R^n$, therefore for all $(u,v)\in (\Bbb R^n)^\prime\times \Bbb R^n\equiv \Bbb R^n\times \Bbb R^n$
$$
\langle u, \lambda v\rangle = \lambda \langle u, v\rangle\quad\forall\lambda\in\Bbb R^n.\label{1}\tag{SP}
$$
Now, if $y$ is any vector contained in a spherical neighborhood $V_x$ of $x$ (with $v_x\Subset K$), we have by hypothesis $\langle F(x), y-x \rangle\ge 0$  then $\langle F(x), x-y \rangle=-\langle F(x), y-x \rangle$ is $\le 0$ by \eqref{1}. However, by hypothesis we have that 
$$
\langle F(x), x-y \rangle \ge 0\:\text{ again}\iff \langle F(x), y-x \rangle=0 \quad\forall y\in V_x
$$
and this implies $F(x)=0$.
