Proof of Nearest Points in a Nonempty Closed Convex Subset of $\mathbb{R}^n$: Existence, Uniqueness and the Variational Inequality

Throughout, closedness and compactness in $\mathbb{R}^{n}$ refer to the metric topology $\mathcal{T}_{d_{E}}$, which exists by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii), and a point $p\in C$ is called nearest if $\lVert x-p\rVert\le\lVert x-y\rVert$ for every $y\in C$.

We first record an expansion. For $p,y\in\mathbb{R}^{n}$ and $t\in\mathbb{R}$ let $z_{t}=(1-t)p+ty=p+t(y-p)$. Since $x-z_{t}$ is the sum of $x-p$ and $-t(y-p)$, Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (viii), applied to these two vectors, together with bilinearity of the dot product gives

$$\lVert x-z_{t}\rVert^{2}=\lVert x-p\rVert^{2}-2t\,(x-p)\cdot(y-p)+t^{2}\lVert y-p\rVert^{2}.$$

We now prove (ii). Let $p\in C$. Suppose first that $(x-p)\cdot(y-p)\le0$ for every $y\in C$, and let $y\in C$. The expansion with $t=1$, for which $z_{1}=y$, gives

$$\lVert x-y\rVert^{2}=\lVert x-p\rVert^{2}-2\,(x-p)\cdot(y-p)+\lVert y-p\rVert^{2}\ge\lVert x-p\rVert^{2},$$

and since both norms are nonnegative, $\lVert x-y\rVert\ge\lVert x-p\rVert$. So $p$ is nearest. Conversely, let $p$ be nearest, let $y\in C$, put $s=(x-p)\cdot(y-p)$, and suppose that $s>0$. Then $y\neq p$, so $\lVert y-p\rVert>0$ and we may put

$$t=\min\bigl\{1,\,s/\lVert y-p\rVert^{2}\bigr\}\in(0,1].$$

The point $z_{t}$ lies on the segment $[p,y]$, which is contained in $C$ because $C$ is convex (Segments and Convex Subsets of $\mathbb{R}^n$ (i), Segments and Convex Subsets of $\mathbb{R}^n$ (ii)). As $p$ is nearest, $\lVert x-z_{t}\rVert^{2}\ge\lVert x-p\rVert^{2}$, and the expansion together with $t\lVert y-p\rVert^{2}\le s$ gives

$$0\le\lVert x-z_{t}\rVert^{2}-\lVert x-p\rVert^{2}=t\bigl(t\lVert y-p\rVert^{2}-2s\bigr)\le t(s-2s)=-ts<0,$$

a contradiction. Hence $s\le0$, and (ii) is proved.

Next we show that a nearest point exists. As $C$ is nonempty, fix $c_{0}\in C$, let $r=\lVert x-c_{0}\rVert$, and let $\bar{B}=\bar{B}_{d_{E}}(x,r)$. The closed ball $\bar{B}$ is compact by the standing setting, hence closed by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (ii). So $K=C\cap\bar{B}$ is closed, as the intersection of two closed sets (its complement is the union of their open complements, which is open by Topological Space (iii)); since $K\subseteq\bar{B}$, it is compact by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i). It contains $c_{0}$, since $d_{E}(x,c_{0})=r$ (Closed Ball in a Metric Space (ii)). Consider the function $f:K\to\mathbb{R}$, $f(y)=\lVert x-y\rVert$. By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vii),

$$|f(y)-f(y')|\le\lVert y-y'\rVert=d_{E}(y,y')\qquad(y,y'\in K),$$

so for every $\varepsilon>0$ the choice $\delta=\varepsilon$ meets Continuous Map Between Metric Spaces (i) at each point of $K$, and $f$ is continuous on $K$. By Extreme Value Theorem on a Compact Subset of a Metric Space (i) there is $p\in K$ with $f(p)\le f(y)$ for every $y\in K$. If $y\in C\setminus K$, then $y\notin\bar{B}$, so

$$f(y)=d_{E}(x,y)>r=f(c_{0})\ge f(p).$$

Hence $f(p)\le f(y)$ for every $y\in C$, that is, $p$ is nearest.

Finally, a nearest point is unique. Let $p,q\in C$ both be nearest. By (ii), applied to $p$ with $y=q$ and to $q$ with $y=p$, we have $(x-p)\cdot(q-p)\le0$ and $(x-q)\cdot(p-q)\le0$. Adding these and using bilinearity,

$$\lVert q-p\rVert^{2}=\bigl((x-p)-(x-q)\bigr)\cdot(q-p)=(x-p)\cdot(q-p)+(x-q)\cdot(p-q)\le0,$$

so $\lVert q-p\rVert=0$ and $q=p$. Together with the existence shown above, this proves (i).