How can I prove that $f(x)$ is convex on an interval if and only if $f'(x)$ is monotonically increasing ?
• Let $\lambda \in (0,1)$ and $x,y \in I$. $f(x)$ is convex on an interval, if the statement $f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y) $ is true for all $x,y$.
• $f'(x)$ is monotonically increasing on an interval $I$, if $0 \leq \frac{d f'(x)}{dx}$ is true for all $x \in I$.
I don't have a problem understand this, with a graph it's quite easy to understand. But I can't manage to prove it formally, can someone give me a hint?
Note that I do have seen this question, but since I can't use inequality proven previously it doesn't help me.
Best Answer
Hint: Convexity of $f$ can be expressed as $$a < x < b \in I \implies f (x) \leq f(a) + \frac{f(b) - f(a)}{b -a} (x - a)$$
or $$a < x < b \in I \implies f (x) \leq f(b) + \frac{f(b) - f(a)}{b -a} (x - b)$$
Then $f$ is convex in $I$ if, and only if,
$$a < x < b \in I \implies \frac{f (x) - f(a)}{x-a}\leq \frac{f(b) - f(a)}{b -a} \leq \frac{f(b) - f(x)}{b -x}$$
In fact, one needs only one of those inequalities to characterize $f$ convex.
Edit: Let $x \to a$ on the first inequality and $x \to b $ on the second inequality then
$$\begin{align} a < b \implies f'(a) &\leq \frac{f(b) - f(a)}{b -a} \leq f' (b) \\a < b &\implies f'(a) \leq f'(b) \end{align}$$ Can you take it from here?