Problem: Suppose that $$f:\mathbb{R}\rightarrow\mathbb{R}$$ is continuous. Furthermore, suppose that the right derivative exists everywhere and is bounded, i.e., $|f'_{+}|<M$. (or suppose that the left derivative exists everywhere and is bounded). Then $f$ is Lipschitz continuous.
I tried to prove this by all I managed to show is that $f$ is locally Lipschitz continuous.
My Attempt: From the definition of the right derivative we have
\begin{align}
|f'_{+}(x)| =\left|\lim_{h\rightarrow0^+} \frac{f(x+h)-f(x)}{h}\right|\leq M
\end{align}
Fix $\epsilon>0$. Then from the above we know that there exists $h>0$ such that for all $h'<h$ we have
\begin{align}
\left| \frac{f(x+h')-f(x)}{h'}\right|\leq M+\epsilon
\end{align}
It follows that for all $y\in\mathbb{R}$ such that $|y-x|<h$ we have
\begin{align}
\left| \frac{f(y)-f(x)}{y-x}\right|\leq M+\epsilon
\end{align}
So this shows that $f$ is locally Lipschitz continuous. But I cannot see how to extend this approach to the whole domain.
Further context: I am self-studying Royden and a similar result appears in the chapter on absolutely continuous functions (Chapter 5 in 3rd edition) as an exercise and I was not able to crack it either.
Best Answer
Consider $g(x)=f(x)+Mx$; continuous, right differentiable and $g_{+}'(x)>0$. But now for a fixed $x$ there is $h_x>0$ st for $y\in (x, x+h_x)$ we have $g(y)>g(x)$ so $g$ is locally increasing (from the right); a standard argument implies that $g$ is (non necesarily strictly) increasing on the real line
(assume $y>x, g(y) < g(x)$; let $w=\inf_{z\in [x,y]}g(z) < g(x)$; by the above $x \ne w$ and then $g(x)<g(w-\epsilon)$ so $g(x)=g(w)$ but by definition of $w$, there is $z_n \to w,z_n >w, g(z_n)<g(x)=g(w)$ contradicting the existence of $h_w$ above)
So $f(x)+Mx \le f(y)+My, y>x$ so $f(x)-f(y) \le M(y-x)$
Consider $h(x)=f(x)-Mx$; the same reasoning shows that one is (not necessarily strictly) decreasing on the real line, so $f(x)-Mx \ge f(y)-My, y>x$ so $f(y)-f(x) \le M(y-x)$
Together the two inequalities show $|f(x)-f(y)| \le M|x-y|$