Does the Border (Boundary) Points of a convex body make a concave function?

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.

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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

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 $1$.