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?! 
What's your opinion on this?
 A: 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.
A: 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.
