# Under what conditions does local concavity imply global concavity?

I have the following result:

Assume $U:\mathbb{R}^+\to\mathbb{R}^+$ is continuous and strictly increasing. Further, for every $a>0$ there exists a neighborhood (interval) $S$ of $a$ such that $U$ is concave on $S$.

Now I want to show that $U$ is concave on all of $\mathbb{R}^+$. I was thinking about using the fact that $U$ has a non-increasing right derivative on $int S$ (famous result by Stolz)- by continuity on $S$. However, I don't really know how to incorporate this.

Consider the following result: A function $f:I\to\mathbb{R}$ is concave if and only if for $x,y,z\in I, x<z<y$ it holds $$\frac{f(x)-f(z)}{x-z} \geq \frac{f(x)-f(y)}{x-y} \geq \frac{f(z)-f(y)}{z-y}.$$

My intuition is the following. For any $x<z<y$ in $\mathbb{R}^+$ I know that around each of these points, $U$ is concave. Further, for any points in between $x,z,y$, the same situation holds. Hence, I use overlapping intervals, around the points in between, to show the above inequalities using Stolz result. However, I have the feeling that considered intervals need not necessarily overlap?!

• This is similar to your proposed solution, but do you know that each concave function is (locally) absolutely continuous, where the (almost everywhere existent) derivative is nonincreasing (and vice versa)? – PhoemueX Sep 12 '14 at 21:15

More generally, the result is true in any convex subset $D$ of a topological vector space: if $f: D \to \mathbb R$ is continuous and every $a \in D$ has a neighbourhood on which $f$ is concave then $f$ is concave on $D$.

Proof: Suppose not. Then there exist $a,b \in D$ and $t \in (0,1)$ such that $f(t a + (1-t) b) < t f(a) + (1-t) f(b)$. Define $g: [0,1] \to \mathbb R$ by $g(s) = f(sa + (1-s) b) - s f(a) - (1-s) f(b)$. Then $g$ is continuous, and every $s \in [0,1]$ has a neighbourhood on which $g$ is concave. Note that $g(0) = g(1) = 0$ while $g(t) < 0$. Let $y = \min\{g(s): s \in [0,1]\} < 0$, and $p = \max\{s: g(s) = y\} \in (0,1)$. But $g(s) > g(p)$ on $(p,1]$ while $g(s) \ge g(p)$ on $[0,p)$, so there is no neighbourhood of $p$ on which $g$ is concave.

• +1 I am just not fluent enough in this stuff. I have to read it carefully and slowly, instead of instantly thinking "oh, yes". I guess the problem is I am not currently using it much ... No, that is a feeble excuse! – almagest Sep 12 '14 at 21:24
• The reason that there is no neighborhood of p on which g is concave is that g(p) lies below the secant between any two points in the neighborhood of p? – Tim Sep 12 '14 at 21:42
• Yes (for any two points where one is on one side and the other is on the other). – Robert Israel Sep 13 '14 at 1:37

Just a little addition to Robert Israel's great proof.

I had to think about why every $s\in[0,1]$ has a neighborhood on which $g$ is concave a little and came up with the following reasoning - for those who are interested.

Let $s\in[0,1]$ and consider the point $z(s):=s x + (1-s)y$. By assumption, $f$ is concave on an interval around $z(s)$ whose points can be described by convex combinations of $x$ and $y$. Hence, for $s_1,s_2$ around $s$ such that $$z(s_1) = s_1 x + (1-s_1)y, \quad z(s_2) = s_2 x + (1-s_2)y$$ lie in the interval on which $f$ is concave, by midpoint concavity the following holds \begin{align*} f\left(\frac{1}{2} z(s_1) + \frac{1}{2} z(s_2)\right) &\geq \frac{1}{2} f(z(s_1)) + \frac{1}{2} f(z(s_2))\\ \Leftrightarrow\quad\quad\,\,\, f\left(\frac{1}{2} z(s_1) + \frac{1}{2} z(s_2)\right)&\geq\frac{1}{2} f(s_1 x + (1-s_1)y) + \frac{1}{2} f(s_2 x + (1-s_2)y) \end{align*} \begin{align*} &\Leftrightarrow\quad f\left(\frac{1}{2} z(s_1) + \frac{1}{2} z(s_2)\right) - \left(\frac{1}{2}s_1 + \frac{1}{2}s_2\right) f(x) - \left(1-\frac{1}{2}s_1 - \frac{1}{2}s_2\right) f(y)\geq\\ &\qquad\qquad\qquad\frac{1}{2} f(s_1 x + (1-s_1)y) + \frac{1}{2} f(s_2 x + (1-s_2)y) - \left(\frac{1}{2}s_1 + \frac{1}{2}s_2\right) f(x) \\ &\qquad\qquad\qquad\quad - \left(1-\frac{1}{2}s_1 - \frac{1}{2}s_2\right) f(y)\\ &\Leftrightarrow\qquad\qquad\qquad\, g\left(\frac{s_1+s_2}{2}\right) \geq \frac{1}{2} g(s_1) + \frac{1}{2} g(s_2), \end{align*}
since $$f\left(\frac{1}{2} z(s_1) + \frac{1}{2} z(s_2)\right) = f\left(\left(\frac{1}{2}s_1 + \frac{1}{2}s_2\right)x + \left(1-\frac{1}{2}s_1 - \frac{1}{2}s_2\right) y\right).$$ Hence, for every $s\in[0,1]$ there exists a neighborhood on which $g$ is concave.