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

Question

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!

Answer 1

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 $\lambda = \frac 12$.

Answer 2

May be, you could just use Taylor expansions around $h=0$. This would give $$\text{lhs}=\frac{\varphi(c)-\varphi(c-h)}{h}=\varphi '(c)-\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi ^{(3)}(c)-\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$ $$\text{rhs}= \frac{\varphi(c+h)-\varphi(c)}{h}=\varphi '(c)+\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi ^{(3)}(c)+\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$ $$\text{rhs-lhs}=h \varphi ''(c)+\frac{1}{12} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$

Answer 3

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 $$ \frac{φ(c)-φ(c-h_1)}{h_1}\le k\le \frac{φ(c+h_2)-φ(c)}{h_2}. $$