1
$\begingroup$

Working on some real analysis work, I've been able to show that for a function $f$, which is convex on $[a,b]$, for $a\leq x_1< x_2< x_3\leq b$: $$\frac{f(x_2)-f(x_1)}{x_2-x_1} \leq \frac{f(x_3)-f(x_2)}{x_3-x_2}$$ and for $h>0$, for some $x_0 \in [a,b]$: \begin{equation} \frac{f(x_0+h)-f(x_0)}{h} \end{equation} is non-decreasing.

I'm stuck needing to show that:

1). $\exists \; c_0\in\mathbb{R}$ such that $\frac{f(x_0+h)-f(x_0)}{h} > c_0$ for all $h>0$. Intuition tells me $c_0 = f_+'(x_0)$, but i'm having issues with the proof.

2) $\lim\limits_{h\to 0^+} \frac{f(x_0+h)-f(x_0)}{h}= f_+'(x_0)$

Thanks for any help

$\endgroup$
1
$\begingroup$

Consider some $k > 0$ and any $h > 0$ such that

$$a < x_0-k < x_0 < x_0 + h < b.$$

Using your inequality for a convex function, we have

$$\frac{f(x_0) - f(x_0-k)}{k}= \frac{f(x_0) - f(x_0-k)}{x_0 - (x_0-k)} \leqslant \frac{f(x_0+h) - f(x_0)}{x_0+h - x_0} = \frac{f(x_0+h) - f(x_0)}{h}. $$

Thus, $[f(x_0 + h) - f(x_0)]/h$ is bounded below by $[f(x_0) - f(x_0 -k)]/k.$ Hence, there exists a greatest lower bound $c_0 = \inf\{[f(x_0 + h) - f(x_0)]/h: x_0 < x_0 + h \leqslant b\}$

As you have already shown that $[f(x_0 + h) - f(x_0)]/h$ is non-increasing, we have existence of the limit

$$\lim_{h \to 0+} \frac{f(x_0+h) - f(x_0)}{h} = c_0 := f'_+(x_0).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.