# One-sided derivatives - convexity

Let $$f_{\ell}'\leq f_r'$$ in $$(a,b)$$ and $$\forall x_1 < x_2$$ in $$(a,b)$$ it holds that $$f_r'(x_1)\leq f_{\ell}'(x_2)$$. Then it follows that $$f$$ convex in $$(a,b)$$.

$$f_{\ell}', f_r'$$ are the one-sided derivatives.

I found a hint in the notes for the proof of this statement:

Generalize Rolle's Theorem and then IVT as follows:

Let $$f$$ be continuous on $$[a,b]$$ and the one-sided derivatives exist for each $$x\in (a,b)$$.

Then

(i) If $$f(a)=f(b)$$ then there is a $$\xi\in (a,b)$$ such that $$f_{\ell}'(\xi)f_r'(\xi)\leq 0$$

(ii) $$\exists \xi\in (a,b)$$ : $$\lambda_1(a,b)$$ is between $$f_{\ell}'(\xi)$$ and $$f_r'(\xi)$$.

Then apply the following proposition:

If $$f$$is defined in an interval $$I$$, then $$(a)\iff (b)\iff (c)$$:

(a) $$f$$ is convex

(b) $$\lambda_1$$ is increasing as for $$x_1$$ and $$x_2$$

(c) $$\lambda_2\geq 0$$

It holds that $$\lambda_2=\frac{\frac{f(x_1)-f(x_3)}{x_1-x_3}-\frac{f(x_3)-f(x_2)}{x_3-x_2}}{x_1-x_2}$$



Could you explain to me how we could use that? I haven't really understood that.

First, it follows that $$f$$ is continuous on $$(a,b)$$. Can you see why?
Next, we need a version of the mean value theorem which applies to functions with one-sided derivatives. Let $$f$$ be continuous on the interval $$[p,q]$$ and have one-sided derivatives in the interior $$(p,q)$$. Then there exists a number $$r\in(p,q)$$ such that $$\frac{f(q)-f(p)}{q-p}$$ lies between $$f'_-(r)$$ and $$f'_+(r)$$. (I'm using $$f'_-$$ and $$f'_+$$ instead of $$f'_\ell$$ and $$f'_r$$.) The proof is similar to that of the standard mean value theorem. Define $$F(x)=f(x)-\frac{f(q)-f(p)}{q-p}x$$ $$F$$ is continuous on $$[p,q]$$ and it has one-sided derivatives $$F'_\pm(x)=f'_\pm(x)-\frac{f(q)-f(p)}{q-p}$$ and $$F(p)=F(q)$$. Now apply a generalization of Rolle's theorem to conclude.
Suppose $$x_1 are in the interval $$(a,b)$$. Using the theorem above, it follows that there exist $$c\in(x_1,x_2)$$ and $$d\in(x_2,x_3)$$ such that $$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ lies between $$f'_-(c)$$ and $$f'_+(c)$$, and $$\frac{f(x_3)-f(x_2)}{x_3-x_2}$$ lies between $$f'_-(d)$$ and $$f'_+(d)$$. Using the given hypotheses (note $$c), it follows that $$f'_-(c)\leq \frac{f(x_2)-f(x_1)}{x_2-x_1}\leq f'_+(c)\leq f'_-(d)\leq \frac{f(x_3)-f(x_2)}{x_3-x_2} \leq f'_+(d) \\ \Rightarrow \frac{f(x_2)-f(x_1)}{x_2-x_1}\leq \frac{f(x_3)-f(x_2)}{x_3-x_2}$$ The last inequality is an equivalent definition of convexity in an interval (see here, for instance).