A concave positive function on $[1,\infty)$ is uniformly continuous Let $f$ be a concave positive function on $[1,\infty)$, then $f$ is uniformly continuous on $[1,\infty)$.
This was a true or false problem that I couldn't prove to be true, so I'm thinking that maybe there is a counterexample. I know that for $f$ to be concave then $f''(x)\lt 0$ on $[1,\infty)$.
Does $f(x)=\frac{1}{x-2}$ work as a counterexample? $f''(x)=-1\lt 0$ and at $x=2$ it is undefined so it wouldn't be uniformly continuous on the interval $[1,\infty)$, right?
Uniformly Continuous: Let $E$ be a nonempty subset of $\mathbb R$ and $f:E\to \mathbb R$. The $f$ is uniformly continous on $E$ if and only if for every $\epsilon \gt 0$ there is a $\delta\gt 0$ such that $|x-a|\lt\delta$ and $x,a\in E$ imply $|f(x)-f(a)|\lt \epsilon$.
 A: A hint provided that $f$ is of class $C^1$: assume without loss of generality that $f'(1)$ exists (otherwise replace $1$ with any greater number and recall that continuity on compact subsets implies uniform continuity), and pick $1 \leq x<y$. Then, since $f'$ is decreasing,
$$
f(x)-f(y) = \int_x^y f'(t) \, dt \leq f'(1) (x-y).
$$
Exchanging the rôle of $x$ and $y$,
$$|f(x)-f(y)| \leq |f'(1)| |x-y|,$$ 
and you conclude easily that $f$ is uniformly continuous. I was thinking of a proof under the mere assumption of concavity, without any further regularity, but it looks a bit harder.
A: Let us assume that $f$ is continuous (otherwise it is wrong, obviously).
As $f$ is concave, $f$ has a (maybe infinite) right and left derivative at every point (let us use the notation $f'_+ / f'_-$ for these). Plus, this derivative is decreasing.
In particular, $x>2\implies f'_\pm(x) \le f'_+(2) $. As $f$ is positive, $f'_\pm$ remains positive because
$$
0\le f(x+h) \le f'_\pm(x)h + f(x)
$$
(take $h$ big enough).
This proves that the derivative is bounded on the interval $(2,\infty)$.
On the compact interval $[1,3]$ f is uniformly continuous because it is continuous.
Hence $f$ is uniformly continuous.
A: First some (counter-)examples.
a) $f$ need not be continuous at $1$. Consider the concave function $f(1)=0$, $f(x)=1$, $x>1$.
b) Assuming that $f$ is continuous at $1$, it may still happen that the derivative at $1$ does not exist. Consider the concave function
$$
f(x)=\begin{cases}\sqrt{1-(x-2)^2},&\mbox{if }1\leq x\leq 2\\ 1&\mbox{if } 2\leq x\end{cases}.
$$
Note that $f$ is smooth in $(1,\infty)$. Observe also that the derivative of $f$ is not bounded.
However, the result is still true if you assume that $f$ is continuous at $1$. You can circumvent the use of derivatives in the following way.
Recall that a function is concave if and only if the following inequalities hold for all $x_1<x_2<x_3$
$$
\frac{f(x_2)-f(x_1)}{x_2-x_1}\geq \frac{f(x_3)-f(x_1)}{x_3-x_1}\geq \frac{f(x_3)-f(x_2)}{x_3-x_2}.
$$
1) $f$ is non-decreasing. Assume that there is some $x_1<x_2$ such that $f(x_1)>f(x_2)$. Then we have, for any $x_3>x_2$,
$$
f(x_3)\leq \frac{(f(x_2)-f(x_1))(x_3-x_2)}{x_2-x_1}+f(x_2).
$$
By picking $x_3$ large enough, we see that $f(x_3)<0$, which is a contradiction.
2) $f$ is uniformly continuous in $[1+\epsilon,\infty]$, for all $\epsilon>0$. For any $1+\epsilon\leq x_2<x_3$, we have
$$
\frac{f(x_3)-f(x_2)}{x_3-x_2}\leq \frac{f(x_3)-f(1)}{x_3-1}\leq \frac{f(1+\epsilon/2)-f(1)}{\epsilon/2}.
$$
3) $f$ is uniformly continuous. This is clear since $f$ is continuous in $[1,\infty]$ by 2) combined with the assumed continuity at $1$. Thus, $f$ is uniformly continuous in $[1,3]$. But invoking 2) again, $f$ is also uniformly continuous in $[2,\infty]$.
A: The simplest counter example for this question is $f(x)=x^2$. This function is concave up on $[1,\infty]$, but most real analysis textbooks will show that it is not uniformly continuous. 
