# Show that the set of points that are nearer $a$ than $b$ with respect to $\lVert \cdot \rVert_2$ is convex

I am trying to show the above statement:

Show that the set of points that are nearer $a$ than $b$ in the sense of Euclidean $\lVert\cdot\rVert_2$ are convex.

My attempt

This follows from the Proposition: The convex hull of $M \subset \mathbf R^2$ is convex, indeed the smallest convex set that contains $M$. Moreover, co($M$) is the set of all convex combinations of points of $M$,

$$\text{co}(M) := \{x \in \mathbf R^2 | \exists x_1, x_2, ... , x_N \in M \text{ such that } x = \sum\limits_{i=1}^{N} \lambda_i x_i,\\ \text{ for some } \lambda_1, \lambda_2, ..., \lambda_N, \text{ where } \lambda_1 \geq 0 \forall i, \sum\limits_{i=1}^{N} \lambda_i = 1 \}.$$

The main idea is based on the idea about the corollary 1.14 in http://www.math.udel.edu/~angell/ch1.pdf:

The convex hull of a compact set in $\mathbf R^n$ is compact.

Assume $C \subset \mathbf R^n$ be compact. See the original source for the skipped steps. The sequence $\{A\}_{k=1}^{\infty}$ has a subsequence $\{AA\}_{j=1}^{\infty}$ which converges to a point of co $(C)$ which shows that this latter set is compact.

If $C$ is closed and convex, it has a smallest/nearest element in a certain sense. It is a simple result for analysis that involves the fact that the function $x \to \lVert x\rVert$ is a continuous map from $\mathbf R^n \to \mathbf R$ and the fact that Cauchy sequences in $\mathbf R^n$ converges.

We need also the theorem: Every closed convex subset of $\mathbf R^n$ has a unique element of minimum norm. See the proof in the source above.

There must be also a simpler way to prove the statement. Is there any sense in my attempt? How can you show the statement?

The set is a hyperplane. I think you are over-thinking the problem.

The set in question is $$\left||a-x\right|| \lt \left ||b-x\right|| \Leftrightarrow\left||a-x\right||^2 \lt \left ||b-x\right||^2$$ Writing out the norms we get $$(a\cdot a) - 2 (a \cdot x) + (x \cdot x) < (b\cdot b) - 2 (b \cdot x) + (x \cdot x)$$ Cancelling $(x \cdot x)$ and rearranging $$(c \cdot x) < (b\cdot b) - (a\cdot a) ~\text{ where c = 2(b-a)}$$

Let us write the above inequality as $$(c \cdot x) < d$$

This is clearly a convex set: If $x$ and $y$ are in the set then $$(c \cdot (\lambda x + (1-\lambda) y ) = \lambda (c \cdot x) + (1-\lambda) (c \cdot y)< \lambda d + (1-\lambda) d = d$$

• Not a hyperplane, but a halfspace. Feb 6 '14 at 3:01

Let $\phi(x) = \|x-a\|^2-\|x-b\|^2= \|a\|^2-\|b\|^2+2 \langle b-a, x \rangle$. $\phi$ is affine (that is, linear plus a constant) hence convex, and so $\{ x | \phi(x) < 0\}$ is convex.

Note: Since it also follows from the above that $\{ x | \phi(x) > 0\}$ is convex, we see that the sets are open half-spaces separated by the hyperplane $\{ x | \phi(x) = 0\}$.

• What do you mean by this $\{x|\Phi(x) < 0\}$? Jan 19 '14 at 13:10
• It means the set of $x$ that satisfy the inequality $\phi(x) <0$, where $\phi$ is the function defined on the first line. Jan 19 '14 at 17:07