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I've had a hard time proving this statement. The objective is to prove that the set $M$ is convex where $f(y)$ can be any function. The task is to prove it using triangle inequality. I've looked at threads like Proving Convexity of an Open Disk but i still can't wrap my head around it. If anyone could give me a hint in the right direction i would be very grateful.

$$M = \{\, x\,\big| \, ||x-y|| \leq f(y) \, \text{ for all } y \in S\,\} \quad \text{where } S \subseteq \mathbb{R}^n$$

Best Regards

stressed out
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Pontus S
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2 Answers2

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Hint: If $x_1, x_2 \in M$ and $0 < \lambda < 1$ then for all $y \in S$ $$ |\lambda x_1 + (1-\lambda) x_2 - y| = |\lambda(x_1-y) + (1-\lambda)(x_2 - y)| \, . $$ Now use the triangle inequality to conclude that this is $$ \le \lambda| x_1-y| + (1-\lambda)|x_2 - y| \le \lambda f(y) + (1-\lambda) f(y) = f(y) \, , $$ which means that $\lambda x_1 + (1-\lambda) x_2 \in M$.

Martin R
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  • Seems to me that one can conclude that $|x_1-y|$ and $|x_2-y|$are both $\leq f(y)$ as that's stated in the original formula. Then one should be able to rewrite it all:

    $|\lambda( x_1 - y) + (1- \lambda) (x_2 - y)| \leq \lambda |x_1-y| + (\lambda - 1) |x_2-y| \leq f(y)$

     – Pontus S Feb 21 '19 at 14:05
  • @PontusS: Correct. – Martin R Feb 21 '19 at 14:10
  • A short final question to make sure i understand this correctly. For a convex set a line between two points in the set must be contained within the set. Two arbitrary points $x_1$ and $x_2$ are contained in the set.

    The equation

    $\lambda | x_1 - y| + (1- \lambda) |x_2 -y| \leq f(y)$ Proves that all the points on the line connecting the points $x_1$ and $x_2$ are also contained within the set.

    Sorry for any inconvenience just want to make sure i get this stuff :) You have been very helpful thanks!

     – Pontus S Feb 21 '19 at 14:34
  • @PontusS: $|(\lambda x_1 + (1-\lambda) x_2) - y| \le \ldots \le f(y)$ for all $y \in S$ shows that $\lambda x_1 + (1-\lambda) x_2$ is in the set. – Martin R Feb 21 '19 at 14:38
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Another hint: $M$ is the intersection of a certain collection of closed balls.

kimchi lover
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