# 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.

1. I don't see any translation to the origin or anything, so this confused me, are they assuming $$x$$ is the origin or something?
2. How does $$F(x) = 0$$ follow from the inequality.

1. 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.

1. 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 $$F(x)=0$$.

• How is $\langle F(x), x-y \rangle$ greater than $0$? My hypothesis is with $y-x$ right? – yoshi Feb 23 at 22:48
• Yes, you hypothesis is that $\langle F(x), y-x\rangle\ge 0$ for all $y\in K$: therefore if you choose $y$ sufficiently close to $x$, for example requiring $y\in V_x \Subset K$, you can find $y^\prime\in V_x$ such that $$y^\prime -x =-(y-x)=x-y\iff y^\prime=x+(x-y)$$ and the non negativity condition holds. – Daniele Tampieri Feb 24 at 10:01
• Okay, Now I understand the statement $\langle F(x), y-x \rangle = 0$ for all $y \in V_x$. Why does it follow that $F(x) = 0$? The condition just gives that $F(x)$ and $y-x$ are orthogonal in $V_x$. – yoshi Feb 24 at 14:36
• o wait: okay: math.stackexchange.com/questions/400331/… – yoshi Feb 24 at 14:40
• Yes, you have found the answer by yourself. Thank you for accepting my answer. – Daniele Tampieri Feb 24 at 14:45