Continuing were we left on a previous post about non differentiable functions that however have a well defined right derivative, we want now to show that if the function is also continuous, then from the sign of the right derivative we can draw conclusions on the monotonicity of the function.

Theorem. Let $$f(x)$$ be a continuous function over the interval $$[a,b]$$ for which it is defined the right derivative

$f’_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}$

in $$[a,b)$$. Suppose also that $$f’_+(x)\geq 0$$ for all $$x\in[a,b)$$. Then for all $$a\leq x\leq y\leq b$$ we have $$f(x) \leq f(y)$$.

We follow the approach of the Mean Value Theorem’s demonstration. Therefore we first need a generalization of Rolle’s Theorem.

Lemma. If $$f(x)$$ is a continuous function in $$[a,b]$$ that has right derivative $$f’_+(x)$$ in $$[a,b)$$, and such that $$f(a) = f(b)$$, then there exist points $$\alpha$$ and $$\beta$$ in $$[a,b)$$ such that $$f’_+(\alpha)\geq 0$$ and $$f’_+(\beta)\leq 0$$.

Since $$f(x)$$ is continuous, by Weierstrass’s Theorem it must reach its maximum and minimum in $$[a,b]$$. If $$f(a) = f(b)$$ is both maximum and minimum, then the function is constant and the assertion is proved. If $$f(a) = f(b)$$ is a minimum, then the maximum must be a $$\beta \in (a,b)$$, and we have $$f(x) \leq f(\beta)$$ for all $$x \in [a,b]$$. As a consequence, if $$x\in (\beta,b]$$, then $$\frac{f(x)-f(\beta)}{x-\beta}\leq 0$$ and thus $$f’_+(\beta)\leq 0$$. Furthermore, by minimality of $$a$$, we have $$\frac{f(x)-f(a)}{x-a}\geq 0$$ for all $$x\in(a,b)$$, so that $$f’_+(a)\geq 0$$; and we can take $$\alpha = a$$. The demonstration follows the same path in the case where $$f(a) = f(b)$$ is a maximum, and where both maximum and minimum lie in the open interval $$(a,b)$$. $$\square$$

We are ready now to demonstrate the Theorem by contradiction. Suppose there exist $$c_1,c_2\in[a,b]$$, with $$c_1<c_2$$ such that $$f(c_1) > f(c_2)$$, and define the auxiliary function

$F(x) = f(c_2)-f(x)+K(x-c_2)$

with

$K=\frac{f(c_2)-f(c_1)}{c_2-c_1}< 0.$

$$F(x)$$ satisfies the hypotheses of the Lemma in $$[c_1,c_2]$$, since we have $$F(c_1) = F(c_2) = 0$$ and a well defined right derivative, given by

\begin{eqnarray}\lim_{h\rightarrow 0^+} &=& \frac{F(x+h)-F(x)}{h}=\\ &=&\lim_{h\rightarrow 0^+} \frac{-f(x+h)+f(x)+Kh}{h}=\\ &=& -f’_+(x) + K.\tag{1}\label{eq607:1}\end{eqnarray}

Consequently, there must exist a point $$c\in[c_1,c_2)$$ such that

$F’_+(c) \geq 0,$

which, with \eqref{eq607:1}, implies

$f’_+(c) \leq K<0,$

in contradiction with the hypothesis. $$\blacksquare$$