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. \(\blacksquare\)