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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.
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  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 $F(x)=0$.

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  • $\begingroup$ How is $\langle F(x), x-y \rangle$ greater than $0$? My hypothesis is with $y-x$ right? $\endgroup$ – yoshi Feb 23 at 22:48
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    $\begingroup$ 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. $\endgroup$ – Daniele Tampieri Feb 24 at 10:01
  • $\begingroup$ 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 $V_x$. $\endgroup$ – yoshi Feb 24 at 14:36
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    $\begingroup$ o wait: okay: math.stackexchange.com/questions/400331/… $\endgroup$ – yoshi Feb 24 at 14:40
  • $\begingroup$ Yes, you have found the answer by yourself. Thank you for accepting my answer. $\endgroup$ – Daniele Tampieri Feb 24 at 14:45

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