# $A$ convex subset of a set has 'smaller' boundary than the set?

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 $B$.

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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. :) –  Babak S. 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.

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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 $B$.

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@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