The definition I have of a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ is that for every $x, y \in \mathbb{R}$ and every $\lambda \in [0, 1]$,

$$

f(\lambda x + (1-\lambda )y) \leq \lambda f(x) + (1- \lambda )f(y).$$

By proving that slopes increase I mean that for $x \leq y \leq z$, we get $$\frac{f(y) – f(x)}{y-x} \leq \frac{f(z) – f(x)}{z-x} \leq \frac{f(z) – f(y)}{z-y}. $$

Is there a simple proof of this which doesn't assume that such a convex function has a non-negative second derivative? It's difficult to see how the definition gets us here.

## Best Answer

Hint.Note that if $x\leqslant y\leqslant z$, then we can write $$ y=\lambda x+(1-\lambda)z $$ for some $\lambda\in[0,1]$.

Explicitly, you can find what is $\lambda$ in terms of $x,y$ and $z$. On the other hand, the definition of convexity tells you $$ f(y)\leqslant \lambda f(x)+(1-\lambda)f(z). $$ Now do some algebra to get what you want.