if a set S is convex then $\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $ for any z in the set

Question

let S be a convex set, and let $x\notin S;y,z\in S$. such y is the closest to x in S. prove: $\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $. The inner product here is of course the euclidean one. and the distance is with the regular Euclidean distance. I tried using Cauchy Schwartz, but it has not been successful.

Answer 1

You want to show that $$ \langle x-y,\frac{x}{2}+\frac{y}{2}\rangle - \langle x-y,z\rangle = \langle x-y,\frac{x}{2}-\frac{y}{2}+y-z\rangle \\ = \frac 12\Vert x-y \Vert^2 + \langle x-y, y-z \rangle $$ is strictly positive. It suffices to show that $$ \tag {*} \langle x-y, y-z \rangle \ge 0 $$ and that is true because $y$ is the closest point to $x$ in $S$: The function $$ [0, 1] \ni t \mapsto f(t) = \Vert x- (y+t(z-y))\Vert^2 \\= \Vert x-y \Vert^2 -2 t \langle x-y, z-y \rangle + t^2 \Vert z-y \Vert^2 $$ is minimal at $t=0$, therefore $$ f'(0) = -2 \langle x-y, z-y \rangle \ge 0 $$ and that proves $(*)$.

Answer 2

Here is geometric proof. Let $t={1\over 2}(x+y).$ (We shall only use that $t$ is strictly between $x$ and $y$ on the line segment $xy.$) Let $B$ be the closed ball of radius |x-y| and center $x.$ Observe that $S\cap B=\{y\}.$ Let $H$ be the hyperplane through $y$ with normal $x-y.$ If $z$ were on the same side of $H$ as $t,$ then the line segment $yz$ would intersect $B$ at points other than $y,$ a contradiction. Thus, $z$ is on $H$ or on the opposite side from $t.$ Therefore, $\langle x-y,t-z\rangle\gt 0.$ In other words, $$\left\langle x-y,z\right\rangle <\left\langle x-y,t\right\rangle.$$