# A real differentiable function is convex if and only if its derivative is monotonically increasing

I'm working on a problem in baby Rudin, Chapter 5 Exercise 14 reads:

Let $f$ be a differentiable real function defined in $(a,b)$. Prove that $f$ is convex if and only if $f'$ is monotonically increasing.

I am trying to prove that $f'$ is monotonically increasing under that assumption that $f$ is convex. I have written a proof, but a friend and I do not agree on the validity of my argument. Here is my argument.

First, assume that $f$ is convex. Since $f$ is differentiable and real on $(a,b)$, $f$ is continuous on $(a,b)$. So, for $a<s<u<v<t<b$, by the mean value theorem there exist points $y_1\in[s,u]$ and $y_2\in[v,t]$ such that $$f'(y_1)=\frac{f(u)-f(s)}{u-s}\quad\text{and}\quad f'(y_2)=\frac{f(t)-f(v)}{t-v}.$$

Then, by exercise 23 of chapter 4 (proven previously): $$\frac{f(u)-f(s)}{u-s}\leq\frac{f(t)-f(v)}{t-v}$$ or, $$f'(y_1)\leq f'(y_2).$$ Hence, $f'$ is monotonically increasing. $\hspace{3.5in}\square$

My friend claims that this merely proves that for any two arbitrary intervals $[s,u]$ and $[v,t]$ there are points that satisfy $f'(y_1) \leq f'(y_2)$. He says that what I need to prove is that for any two arbitrary points $y_1\leq y_2$ we have $f'(y_1)\leq f'(y_2)$. (I should mention that he doesn't know how one would do this.)

Is he right? Is my proof insufficient? If so, how can I fix it?

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Your friend is correct. If you want to show that a function is monotonically increasing you have to choose $y_1 < y_2$ and then show $f'(y_1) \leq f'(y_2)$. You don't choose $y_1$ and $y_2$; they come as a consequence of your choice of $s,u,v,t$. There's no guarantee that every $y_1$ and $y_2$ in $(a,b)$ can be gotten by choosing $s,u,v,t$ as you've done. – Dimitrije Kostic Nov 14 '11 at 4:25
+1 for showing us your work and asking a clear question. – Srivatsan Nov 14 '11 at 4:27

From the previously solved exercise, you can show that for arbitrary $s<t$ in $(a,b)$, $\lim\limits_{u\to s+}\frac{f(u)-f(s)}{u-s}\leq\lim\limits_{v\to t-}\frac{f(t)-f(v)}{t-v}$, so $f'(s)\leq f'(t)$.