Suppose $B=\{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 + z^2 \le 1\}$.
Consider a point $v \notin B$. How do I prove that there exist a unique point $v_0 \in B$ that is closest to $v$. Also, how do I obtain a formula for this point $v_0$?
I've proved earlier that $B$ is a convex set.
I've considered orthogonal projection onto subspace and alike, but $B$ is not a subspace so my idea fails. Intuitively, such a point $v_0$ must exist, but how do I make the idea rigorous ?
On
Hint: Well you can consider distance function $V = (V_{x},V_{y},V_{z})$, then distance between $V$ and the point of $B$ is $\sqrt{(V_{X}-x)^2+(V_{y}-y)^2+(V_{z}-z)^2}$. And think of the domain you evaluate this function certainly, because of nature of $B$ the domain is closed and bounded. Therefore, by extreme value theorem you can deduce maximum and minimum exist.
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By hyphotesis $B$ is compact; so you can define a function $F:B\longrightarrow \mathbb{R}$ such that maps every $v_0 \in B$ to its distance to $v$. Then by Weierstraß theorem the function $F$ admits minimum and maximum since $B$ is compact, then you can show that this function assume the minimum in only one point. You can see it supposing that there exists two different points $v_0$ and $v_1$ such that $F(v_0)=F(v_1)$ and show that $v_0=v_1$.
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The underlying problem can be formulated as \begin{eqnarray} \text{minimize }\frac{1}{2}\|v_0-v\|_2^2\hspace{1em}\text{ subject to }v_0 \in B. \end{eqnarray}
You are minimizing a $1$-strongly convex function on a closed convex set. The minimizer is unique.
On
Given an arbitrary point $a$ in $\mathbb{R}^3$, the distance from $a$ to the set $B$ is defined by $$\tag{1} d(a,B)=\inf_{b\in B}\|a-b\|. $$ Since the map $$ d^a:\mathbb{R}^3\to [0,\infty), \, d^a(x)=\|a-x\| $$ is (uniformly) continuous, and the set $B$ is compact, there exists some point $p(a)\in B$ such that $$\tag{2} d(a,B)=\|a-p(a)\|. $$ This also means that the infimum in (1) can be replaced by a minimum.
Claim 1: If $a\in \mathbb{R}^3\setminus B$ then $d(a,B)=d(a,\partial B)$.
Proof: Since $B$ is closed, we have $\partial B\subset B$, and thanks to (1), we have $$ d(a,B)\le d(a,\partial B). $$ Also, for every $x,y\in \mathbb{R}^3$ we have: $$ \|a-x\|\le \|a-y\|+\|y-x\|. $$ It follows that $$ d(a,\partial B)=\min_{x\in \partial B}\|a-x\|\le \|a-y\|+\min_{x \in \partial B}\|y-x\| \quad \forall y\in B. $$ Hence $$ d(a,\partial B)=\min_{x\in \partial B}\|a-x\|\le \min_{y\in B}\|a-y\|+\min_{x \in \partial B, y\in B}\|y-x\|, $$ i.e. $$ d(a,\partial B)\le d(a,B)+\underbrace{\min_{x\in \partial B,y\in B}\|x-y\|}_{=0}=d(a,B). $$ This proves that $d(a,B)=d(a,\partial B)$. It also follows that $p(a)\in \partial B$.
Claim 2: If $a\in \mathbb{R}^3\setminus B$, the point $p(a) \in \partial B$ satisfying (2) is unique.
Proof: Now that we know that $p(a)\in \partial B$, we can reformulate the minimization problem as follows:
Minimize $E_a(x)=\frac12\|a-x\|^2$ subject to $F(x)=\frac{\|x\|^2-1}{2}=0$
Let $\bar{x}$ be a solution of the above minimization problem. Thanks to the method of Lagrange Multipliers, there exists some $\lambda\in \mathbb{R}$ such that $$ \nabla E_a(\bar{x})=\lambda\nabla F(\bar{x}), $$ i.e. $$ \bar{x}-a=\lambda \bar{x}. $$ We deduce that $\lambda=\langle \bar{x}-a,\bar{x}\rangle=1-\langle a,\bar{x}\rangle$ and $$ \bar{x}=a+(1-\langle a,\bar{x}\rangle)\bar{x}. $$ Hence $$\tag{3} \bar{x}=\frac{a}{\langle a,\bar{x}\rangle}. $$ From (3) we deduce that $|\langle a,\bar{x}\rangle|=\|a\|$, and therefore $$ \bar{x}=\pm\frac{a}{\|a\|}. $$ Since $$ \left\|a-\frac{a}{\|a\|}\right\|=\|a\|-1<\|a\|+1=\left\|a+\frac{a}{\|a\|}\right\|, $$ it follows that $$ \bar{x}=p(a)=\frac{a}{\|a\|}. $$ This shows that $p(a)$ is uniquely determined by $a$.
Remark: You can prove that for any $a\in \mathbb{R}^3\setminus B$, the point $-p(a)$ is the only solution of the problem
Maximize $E_a(x)$ subject to $F(x)=0$.
The point $v_0$ is the intersection of the surface of the sphere and the ray that comes from the origin and passes through $v$.
Proof: Consider $B'$ the sphere centered at $v$ with radius $r=d(v,v_0)$. Clearly $v_0\in B\cap B'$. Let $w$ be any other point in space $\Bbb R^3$. If $w$ is in the segment $\overline{0v}$, then $w$ is in $\overline{0v_0}$ and hence not in $B'$, or is in $\overline{v_0v}$, and hence not in $B$. If not, triangle inequality gives $d(0,w)+d(w,v)>d(0,v)=r+1$, so $d(0,w)\le1$ and $d(w,v)\le r$ can not hold simultaneously.