Prove the set $S_{\mu}$ is convex

Question

Given the set $S_{\mu}=\{x\in\mathbb{R}^n:||x-a||\leq\mu||x-b||\}$ prove for which $\mu$ $S_{\mu}$ is convex $(a\neq b)$.
I'm not sure if my way was good.
let $x,y\in S_{\mu}$ and $0\leq\lambda\leq1$.
$||\lambda x+(1-\lambda)y-a||=||\lambda(x-a)+(1-\lambda)(y-a)||\leq \lambda||x-a||+(1-\lambda)||y-a||\leq\mu(\lambda||x-b||+(1-\lambda)||y-b||)$.
and this is only possible if $\mu>0$.

Answer 1

A warm thank to Shubham Johri who has pointed a big error of mine.

Two parts 1) and 2):

1. Where your argumentation is flawed.

You have to prove that

$$x,y\in S_{\mu} \implies \forall \lambda \in [0,1], x,y\in S_{\mu}$$

$$\iff \ \|\lambda x+(1-\lambda)y-a\|\le\mu\|\lambda x+(1-\lambda)y-b\| \tag{0}$$

What you do is to write the LHS of (0) in this way:

$$||\lambda(x-a)+(1-\lambda)(y-a)||$$

and then use majorations:

$$\leq \lambda||x-a||+(1-\lambda)||y-a||\leq\mu(\lambda||x-b||+(1-\lambda)||y-b||)$$

(nothing false ; but under the condition that $\mu>1$ !?).

But now, you are in the wrong position to apply triangle inequality and get the majoration of the RHS of (0).

2. A proof.

A first remark is that $\mu<0$ corresponds visibly to an impossible case when you see the definition, whereas the locus is surely nonvoid when $\mu \ge 0$ because considering points $x$ close enough to $a$ warrants the inequality.

Result: the convexity occurs for $0 \le \mu \le 1$.

We are going to prove explicitly that the locus is

• a ball if $0<\mu<1$,

• a point is $\mu=0$,

• a hyperplane is $\mu=1$.

Here is why. Let us start from :

$$x \in S_{\mu} \ \iff \ ||x-a||^2\leq\mu^2||x-b||^2 \tag{1}$$

Let us treat the case $n=2$, for the sake of simplicity (the general case is straightforward). (1) can be written with evident notations:

$$(x-a_1)^2+(y-a_2)^2\le \mu^2((x-b_1)^2+(y-b_2)^2)$$

$$(x^2+y^2)(1-\mu^2)-2(a_1-\mu^2b_1)x-2(a_2-\mu^2b_2)y+(a_1^2-\mu^2b_1^2+a_2^2-\mu^2b_2^2) \leq 0\tag{1}$$

Consider now 3 cases.

• If $0 \leq \mu \leq 1$, dividing (1) by $1-\mu^2>0$ gives an inequation of the form:

$$x^2+y^2-2x_cx-2y_cy+k \le 0$$

which is the interior of a disk with center $(x_c,y_c)$. Therefore, we have a convex set.

• If $\mu > 1$, dividing (1) by $1-\mu^2>0$ gives an inequation of the form (take care to the reversal of the inequality sign):

$$x^2+y^2-2x_cx-2y_cy+k \ge 0$$

which is the exterior of a disk with center $(x_c,y_c)$. Therefore, we do not have a convex set.

• if $\mu=1$**: the first term in (1) is cancelled, yielding a first degree equation, which is in a natural manner the perpendicular bissector of line segment $[a,b]$, a convex set.

Remark: the center of the disk is given by:

$$x_c=\dfrac{a_1-\mu^2b_1}{1-\mu^2}, \ \ y_c=\dfrac{b_1-\mu^2b_2}{1-\mu^2}$$

naturally situated on line segment $[a,b]$ (interior of a so-called "apollonian circle").

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Here is a little graphical aid for the case $a=(0,0),b=(6,0),0\le\mu\le3$ (realized by @Shubham Johri):