Q:

Convex Differentiable Non-Increasing Function is smaller than derivative at zero

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable non-increasing convex function, such that $f\left(0\right)=0$. Show that there exists some $x_0>0$ such that for all $0<x<x_0$, we have $$ f\left(x\right)\leq x\frac{1}{2}f'\left(0\right)$$
I have met this claim in some book, and I am not sure how to prove it, because in the Wikipedia page for convex function it seems like the opposite is claimed, namely, that $$ f\left(x\right)\geq f\left(y\right)+f'\left(y\right)\left(x-y\right)\,\,\,\forall\left(x,\,y\right)$$
then applying it with $y=0$ we find $$ f\left(x\right)\geq f'\left(0\right)x $$
What gives? 

A:

Suppose $f'(0) = 0.$ Since $f$ is nonincreasing, $f(x) \le f(0) =0 = (f'(0)/2)x$ for $x \ge 0.$
Suppose $f'(0)<0.$ Consider
$$\tag 1 \frac{f(x) - f(0)}{x-0} = \frac{f(x)}{x}.$$
This converges to $f'(0)$ as $x\to 0.$ Because $f'(0) < f'(0)/2$, there exists $\delta > 0$ such that $(1)< f'(0)/2$ for $0<|x-x_0|<\delta.$ If $x\in (0,\delta),$ then we can multiply by $x$ and preserve the inequality.  I.e., $f(x) < (f'(0)/2)x$ for $x\in (0,\delta).$

