If $f(x)$ is convex on $[a,b]$, show $h(x) = \frac{f(x) - f(a)}{x - a}$ is increasing

Question

I am unsure how to go about starting the proof. I know if $f(x)$ is convex, then for any $x$ we have:

$$f(x) \geq f(a) + f'(a)(x-a)$$

Which rearranged gives us

$$f'(a) \leq \frac{f(x) - f(a)}{x - a} = h(x)$$

I proved a result that said $f'(x)$ is increasing given $f(x)$ is convex, would that be relevant here?

Answer 1

For $b>x>y>a$, we have \begin{align*} f(y)&=f\left(\dfrac{y-a}{x-a}\cdot x+\dfrac{x-y}{x-a}\cdot a\right)\\ &\leq\dfrac{y-a}{x-a}\cdot f(x)+\dfrac{x-y}{x-a}\cdot f(a), \end{align*} then \begin{align*} f(y)-f(a)\leq\dfrac{y-a}{x-a}\cdot f(x)+\dfrac{a-y}{x-a}\cdot f(a), \end{align*} and hence \begin{align*} \dfrac{f(y)-f(a)}{y-a}\leq\dfrac{f(x)-f(a)}{x-a}. \end{align*}

Answer 2

$$ f(a)+\lambda(f(b)-f(a)) \ge f(a+\lambda(b-a))\rightarrow \frac{f(a+\lambda(b-a))-f(a)}{\lambda(b-a)} \le \frac{\lambda(f(b)-f(a))}{\lambda(b-a)} $$

now making $\lambda \rightarrow 0$ we have

$$ f'(a) \le \frac{f(b)-f(a)}{b-a} $$

but choosing now $b = x$ we have $f'(a) \le \frac{f(x)-f(a)}{x-a}$ as expected.

Here $a \le x \le b$