Minimizing a function with a given inner product

Question

I'm trying to solve a question but a not sure on how to approach. Here is the question

Let $V$ be a finite dimensional space with a given inner product and $A \subset V$ be a convex set and $a \notin A $, $a \in V$. Define $f(x) = ||x - a ||$. Prove that exist at maximum one $x$ that minimized f.

I don't know if a have to prove that exist on x because the statement says "Prove that exist at maximum one $x$".

Answer 1

I think we cannot prove the existence since we would need closedness of $A$ and completeness of $V$ in order to prove that. As for existence, if you have the closedness and completeness, here is a proof.

Let $V$ is complete and $A$ is closed. And let \begin{align*} \delta = \inf_{k\in A}\|{a-k}\| \end{align*}

Then, $\exists \{k_n\}\subset K$ such that $\|{a-k_n}\|\rightarrow \delta$. Observe that, by parallelogram law, \begin{align*} \|{k_n-k_m}\|^2=2\|{a-k_n}\|^2+2\|{a-k_m}\|^2-4\|{a-\frac{1}{2}k_n-\frac{1}{2}k_m}\|^2 \end{align*}

Since $A$ is convex, $\frac{1}{2}k_n+\frac{1}{2}k_m\in A$ $\forall n,m\in \mathbb{N}$. Thus, \begin{align*} \|{k_n-k_m}\|^2\leq 2\|{a-k_n}\|^2 + 2\|{a-k_m}\|^2-4\delta \end{align*}

By sending $n,m\rightarrow \infty$, since RHS converges to $0$, \begin{align*} \|{k_n-k_m}\|\rightarrow 0 \end{align*}

Thus, $\{k_n\}$ is a Cauchy sequence. Since $V$ is Hilbert and $A$ is closed subset, $A$ is complete, so, $\exists k\in A$ such that $k_n\rightarrow k$. And naturally, \begin{align*} \|{k}\|=\|{\lim_{n\rightarrow \infty}k_n}\|=\lim_{n\rightarrow \infty}\|{k_n}\|=\delta \end{align*}

As for uniqueness,

If $\exists k'\in A$ such that $\|{k'}\|=\delta$ and $k\neq k'$, note that $$\delta = \|(1/2)(2a-k-k')\|\leq (1/2)\|a-k'\|+(1/2)\|a-k\|=\delta $$, so $\|2a-k-k'\|=2\delta$.

Then, by Paraleogram law,

$$\|k-k'\|^2 = 2\|a-k\|^2+2\|a-k'\|^2-\|2a-k-k'\|=2\delta+2\delta-2\delta=0. $$

Therefore, $k=k'$ by the definition of the norm.

Answer 2

I believe the question is asking you to prove that under an inner product in a vector space, there exists a unique absolute minimum $x$ of f $x\in A$. Where $f$ is a measure of the distance between $x$ and some point $a$ in the complement of $A$.

Suppose you have 2 minima, $x_1$ and $x_2$.

Let $x=x_1+(1-\lambda)(x_2-x_1)$.

$f(x)=\sqrt{x^2+a^2-2\vec{a}\cdot\vec{x}}$

I believe if you assume $(x_1<>x_2)$ you'll find a contradiction that A is convex.

This intuitively makes sense if you consider a polynomial of degree>2. Local extrema occur at critical points. They usually imply a point of inflection at some point between them and those usually show a change in convexity, meaning the areas on neither side of the curve are entirely convex.

Two minima imply a maxima in between which implies a violation of convexity.

Answer 3

Let $d_A(x) = \inf_{a \in A} \|x - a\|$. Consider the open ball $B(x; d_A(x))$, centred at $x$, with radius $d_A(x)$. Note that, for all $y \in B(x; d_A(x))$, we have $$\|y - x\| < \inf_{a \in A} \|x - a\| \le \|y - a\|$$ for all $a \in A$. Thus, $y \notin A$. Hence, if $B[x; d_A(x)]$ (the closed ball) intersects $A$, the intersection must lie on the sphere $S[x; d_A(x)]$. Thus, $$\operatorname{argmin} f = S[x; d_A(x)] \cap A = B[x; d_A(x)] \cap A,$$ which is the intersection of two convex sets, and hence is convex. But, it's also a subset of the sphere $S[x; d_A(x)]$. Since inner product spaces are strictly convex, this set can be at most a singleton.