MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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 $\mathbb{S}$ be a closed and bounded convex body in 2-D with some non-empty intersection with positive quadrant and let it also contain origin. Let $c>0$ be the right-most point on the x-axis such that $(c,y)\in \mathbb{S}$ for some $y$. Define the function \begin{align} f(x)=\max_{(x,y)~\in~\mathbb{S}}y ~~,x\in[0,c] \end{align} Clearly,for a given $x'$, $f(x')$ is the northernmost point in the vertical strip $x=x'$, is $f(x)$ a concave function? (or does it have some nice properties.).

If you look at $f(x)$, it is the pointwise supremum of an affine function. And also, all the examples I can imagine is concave.

share|cite|improve this question
Isn't it an instant corollary to the very definition of a convex set? (also, it can be the whole $[b;c]$ interval, and never mind $0$). – Włodzimierz Holsztyński May 3 '13 at 2:11
up vote 2 down vote accepted

As Wlodzimierz points out, the answer is trivially yes. Maybe it would help to recall that $S$ is compact, so that for each $x$ there exists $y$ such that $(x,y) \in S$ and $f(x) = y$. So if $f(x_1) = y_1$ and $f(x_2) = y_2$ then the point $\frac{1}{2}((x_1, y_1) + (x_2, y_2)) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ belongs to $S$, and hence $f(\frac{x_1 + x_2}{2}) \geq \frac{y_1 + y_2}{2}$. Also it's easy to see that $f$ is continuous.

But it's interesting to note that continuity fails in ${\bf R}^3$. Let $S$ be the convex hull of the circle in the $xy$-plane with center $(1,0,0)$ and radius $1$, and the point $(0,0,1)$. So $S$ is a cone. Now if we set $f(x,y) = {\rm max}_{(x,y,z) \in S} z$ for $(x,y)$ belonging to the disc with center $(1,0)$ and radius $1$, we find that $f(x,y) = 0$ on the boundary circle except at the point $(0,0)$, where it takes the value $1$.

share|cite|improve this answer
@Nik -- you mean, of course, that the continuity fails (while concavity still holds, and in any dimension). (Just a clarification). – Włodzimierz Holsztyński May 3 '13 at 5:46
@Wlodzimierz: yes. I've edited my answer to clarify this. – Nik Weaver May 3 '13 at 7:51
+1, nice. I got the proof. I am not a mathematician and it is not obvious to me as it is to you people. Can you give an intuitive explanation for this? – dineshdileep May 4 '13 at 6:44

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.