Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ & $B$ be subsets of the real plane. Show that if $A$ is convex and is contained in $B$, which is a bounded set, then the length of the border of $A$ is $\leq$ than that of $B$.

share|cite|improve this question
Wrong. If $B$ is the whole plane and $A$ the unit disc, then $B$ has no boundary at all. – Hagen von Eitzen Dec 20 '12 at 10:19
What is the convex set in geometry? I see you didn't use calculus or real analysis tag for the proble. :) – S. Snape Dec 20 '12 at 10:19
@ Hagen von Eitzen.I should have added that set B is bounded. @Babak Sorouh. Ok. – Sgernesto Dec 20 '12 at 10:24
  1. Given any convex closed subset $A$ of a Euclidean space $\mathbb R^n$, we can define the nearest-point projection $p_A\colon\mathbb R^n\to A$ by sending each point $x\in\mathbb R^n$ into $y\in A$ that attains the minimum $\min_{y\in A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.

  2. A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|\le |x-y|$ for all $x,y\in \mathbb R^n$.

  3. It remains to show that $p_A$ maps $\partial B$ onto $\partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).

The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.

share|cite|improve this answer

First, show that for a general half-space $H$, $B\cap H$ has smaller boundary than $B$.

If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say

$A=B\cap(\bigcap_{k=1}^\infty H_k)$

From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.

share|cite|improve this answer
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve. – Sgernesto Dec 20 '12 at 12:43
In any case I think that an analogous statement is true in any dimension. – Sgernesto Dec 20 '12 at 12:44
@Sgernesto: probably, this is the same argument – Martijn Dec 20 '12 at 15:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.