# Convexity implies $\frac{\varphi(c)-\varphi(c-h)}{h} \leq \frac{\varphi(c+h)-\varphi(c)}{h}$?, $h>0$

Suppose $$\varphi$$ is a convex function on the real line. I wonder if the following is true? For $$h>0$$

$$\frac{\varphi(c)-\varphi(c-h)}{h} \leq \frac{\varphi(c+h)-\varphi(c)}{h}$$

This seems like a trivial fact that should follow from convexity if one considers the intuitive meaning of a convex function as having increasing slopes for consecutive points on its graph.

But I got stuck on which numbers to pick and then use in the convex property. Help would be appreciated!

For $$h > 0$$ we have \begin{align} &\frac{\varphi(c)-\varphi(c-h)}{h} \leq \frac{\varphi(c+h)-\varphi(c)}{h} \\ \iff &\varphi(c)-\varphi(c-h) = \varphi(c+h)-\varphi(c) \\ \iff &\varphi(c) \le \frac 12 \varphi(c-h)+ \frac 12 \varphi(c+h) \end{align} which is exactly the convexity condition for the function $$\varphi$$ $$\varphi(\lambda x_1 + (1-\lambda) x_2) \le \lambda\varphi( x_1) + (1-\lambda) \varphi(x_2)$$ with $$x_1 = c-h$$, $$x_2 = c+h$$, and $$\lambda = \frac 12$$.
May be, you could just use Taylor expansions around $$h=0$$. This would give $$\text{lhs}=\frac{\varphi(c)-\varphi(c-h)}{h}=\varphi '(c)-\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi ^{(3)}(c)-\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$ $$\text{rhs}= \frac{\varphi(c+h)-\varphi(c)}{h}=\varphi '(c)+\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi ^{(3)}(c)+\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$ $$\text{rhs-lhs}=h \varphi ''(c)+\frac{1}{12} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$
By convexity, there is a supporting line at $$(c,φ(c))$$ with slope $$k$$, so that $$φ(x)\ge φ(c)+k(x-c)$$ for all $$x$$ in the domain of $$φ$$. This means that for $$h_1,h_2>0$$ you get $$\frac{φ(c)-φ(c-h_1)}{h_1}\le k\le \frac{φ(c+h_2)-φ(c)}{h_2}.$$