If $K$ is a convex set in this space, then every element $x \in X$ has at most one best approximation in $K$.

Question

I would like to prove the following theorem: Let $(X, \langle \cdot, \cdot \rangle)$ be a vector space over $\mathbb{R}$ with an inner product. If $K$ is a convex set in this space, then every element $x \in X$ has at most one best approximation in $K$.

First, we provide the following definitions:

• Let $(X, \langle \cdot, \cdot \rangle)$ be a vector space over $\mathbb{R}$ with an inner product. Take a non-empty set $K$ in this space and fix $x \in X$. An element $y_0 \in K$ is called the best approximation or nearest point from $x$ to $K$ if $\|x - y_0\| = d(x, K)$.

• Additionally, let $(X, \langle \cdot, \cdot \rangle)$ be a vector space over $\mathbb{R}$ with an inner product. A set $K$ in this space is convex if $\lambda x + (1 - \lambda) y \in K$ for all $\lambda \in [0, 1]$ and for all $x, y \in K$.

Now, I want to prove the above theorem using the following steps:

Let $(X, \langle \cdot, \cdot \rangle)$ be a vector space over $\mathbb{R}$ with an inner product. If $K$ is a convex set in this space, then the following statements hold:

1. $P_K(x) = K \cap \overline{K}_{d(x, K)}(x)$ for all $x \in X$, and hence $P_K(x)$ is a convex set.

2. Every non-empty convex subset of a sphere $S_r(x)$, where $x \in X$, has cardinality equal to $1$.

3. $P_K(x) \subseteq S_{d(x, K)}(x)$ for all $x \in X$.

4. $|P_K(x)| \in \{0, 1\}$ for all $x \in X$.

My attempt: I managed to prove 1.

My attempt for 2.: Let $C \subseteq S_r(x)$ be a non-empty convex subset. Suppose $y_1, y_2 \in C$. Then, by the definition of convexity, for any $\lambda \in [0, 1]$, $\lambda y_1 + (1 - \lambda) y_2 \in C$. However, since $y_1, y_2 \in S_r(x)$, we have $\|x - y_1\| = r$ and $\|x - y_2\| = r$. What can I do now? I write down all I got given and now I am just lost. I was thinking of somehow use parallellogram law, but that is just an idea.

I managed to prove 3.

For 4. I did not have any idea on how to even start. With point 4. I of course prove the theorem I want to prove. But how to do it? How can points 1. - 3. help me here I just cannot see.

Thanks in advance for your help!

Note:

• $P_K(x)$ denotes the set of best approximations (nearest points) from $x$ to $K$.
• $d(x, K)$ is the distance from $x$ to the set $K$.
• $\overline{K}_{d(x, K)}(x)$ is the closed ball of radius $d(x, K)$ centered at $x$.
• $\overline{S}_{d(x, K)}(x)$ is a sphere of radius $d(x, K)$ centered at $x$.

Added attempt:

If $K$ is a non-empty convex subset of the sphere $S_r(x), x \in X$, and suppose that $\exists y_1, y_2 \in K$ such that $y_1 \neq y_2$, then from the convexity of the set $K$ it follows that $\lambda y_1 + (1 - \lambda) y_2 \in K$ $\forall \lambda \in [0, 1]$. Since $y_1, y_2 \in K \subseteq S_r(x)$, it follows that $y_1, y_2 \in S_r(x)$, or $\|x - y_1\| = \|x - y_2\| = r$. Therefore, for $z = \mu y_1 + (1 - \mu) y_2 \in K \subseteq S_r(x), \mu \in (0, 1)$, it consequently holds that $\|x - z\| = \|\mu (x - y_1) + (1 - \mu) (x - y_2)\| < |\mu| \|x - y_1\| + |1 - \mu| \|x - y_2\| = \mu \|x - y_1\| + (1 - \mu) \|x - y_2\| = \mu r + (1 - \mu) r = r$, which is a contradiction since then $z \in K_r(x) \nsubseteq S_r(x)$.

Answer 1

For point 2, $ \subseteq _()$ be a non-empty convex subset. Suppose there are two distinct points $x\neq y \in C$, then $\forall \lambda\in (0,1)$, $\|\lambda x + (1-\lambda) y\| < \lambda\|x\| + (1-\lambda) \|y\| = r$, which violates the convexity of $C$. Hence, $C$ can only have 1 point.

Point 4 is straightforward by combining point 1-3. $P_K(x)$ is a convex (point-1) subset (point-3) of $S_{d(K,x)}(x)$. Thus, either it's empty ($|P_K(x)|=0$), or if nonempty then $|P_K(x)|=1$ (point 2).