Let S be a non-empty closed convex set in $mathbb{R}^n$ and let $ynotin S$, then $exists$ a point $bar{x}in S$ with minimum distance from y, i.e.,$left | y-bar{x} right | leq left | y-x right | forall x in S.$

Furthermore, $bar{x}$ is a minimizing point if and only if $left ( y-hat{x} right )^{T}left ( x-hat{x} right )leq 0$ or $left ( y-hat{x}, x-hat{x} right )leq 0$

Proof

Existence of closest point

Since $Sne phi,exists$ a point $hat{x}in S$ such that the minimum distance of S from y is less than or equal to $left | y-hat{x} right |$.

Define $hat{S}=S cap left { x:left | y-x right |leq left | y-hat{x} right | right }$

Since $ hat{S}$ is closed and bounded, and since norm is a continuous function, then by Weierstrass theorem, there exists a minimum point $hat{x} in S$ such that $left | y-hat{x} right |=Infleft { left | y-x right |,x in S right }$

Uniqueness

Suppose $bar{x} in S$ such that $left | y-hat{x} right |=left | y-hat{x} right |= alpha$

Since S is convex, $frac{hat{x}+bar{x}}{2} in S$

But, $left | y-frac{hat{x}-bar{x}}{2} right |leq frac{1}{2}left | y-hat{x} right |+frac{1}{2}left | y-bar{x} right |=alpha$

It can”t be strict inequality because $hat{x}$ is closest to y.

Therefore, $left | y-hat{x} right |=mu left | y-hat{x} right |$, for some $mu$

Now $left | mu right |=1.$ If $mu=-1$, then $left ( y-hat{x} right )=-left ( y-hat{x} right )Rightarrow y=frac{hat{x}+bar{x}}{2} in S$

But $y in S$. Hence contradiction. Thus $mu=1 Rightarrow hat{x}=bar{x}$

Thus, minimizing point is unique.

For the second part of the proof, assume $left ( y-hat{x} right )^{tau }left ( x-bar{x} right )leq 0$ for all $xin S$

Now,

$left | y-x right |^{2}=left | y-hat{x}+ hat{x}-xright |^{2}=left | y-hat{x} right |^{2}+left |hat{x}-x right |^{2}+2left (hat{x}-x right )^{tau }left ( y-hat{x} right )$

$Rightarrow left | y-x right |^{2}geq left | y-hat{x} right |^{2}$ because $left | hat{x}-x right |^{2}geq 0$ and $left ( hat{x}- xright )^{T}left ( y-hat{x} right )geq 0$

Thus, $hat{x}$ is minimizing point.

Conversely, assume $hat{x}$ is minimizimg point.

$Rightarrow left | y-x right |^{2}geq left | y-hat{x} right |^2 forall x in S$

Since S is convex set.

$Rightarrow lambda x+left ( 1-lambda right )hat{x}=hat{x}+lambdaleft ( x-hat{x} right ) in S$ for $x in S$ and $lambda in left ( 0,1 right )$

Now, $left | y-hat{x}-lambdaleft ( x-hat{x} right ) right |^{2}geq left | y-hat{x} right |^2$

And

$left | y-hat{x}-lambdaleft ( x-hat{x} right ) right |^{2}=left | y-hat{x} right |^{2}+lambda^2left | x-hat{x} right |^{2}-2lambdaleft ( y-hat{x} right )^{T}left ( x-hat{x} right )$

$Rightarrow left | y-hat{x} right |^{2}+lambda^{2}left | x-hat{x} right |-2 lambdaleft ( y-hat{x} right )^{T}left ( x-hat{x} right )geq left | y-hat{x} right |^{2}$

$Rightarrow 2 lambdaleft ( y-hat{x} right )^{T}left ( x-hat{x} right )leq lambda^2left | x-hat{x} right |^2$

$Rightarrow left ( y-hat{x} right )^{T}left ( x-hat{x} right )leq 0$

Hence Proved.