If one of the Dini derivatives is bounded, then f is Lipschitz 
If one of its Dini derivatives (say $D^+$) is bounded show that a function $f$ satisfies a Lipschitz condition, 
Definition of the upper right Dini derivative: $$D^{+}f(x) =
 \limsup_{h\to\ 0^+} \frac{f(x+h) - f(x)}{h}$$

This is a question appearing in Royden's Real Analysis textbook. (Edition 3)  
 A: I knew somebody would find a reason to distrust Royden--his text has survived way too long and none of us later textbook writers have been able to knock it out.
Take the function $f(x)=x$ for $-\infty < x < 0$ and $f(x)=1+x$ for $0\leq x < \infty$.  Then $D^+f(x) =1 $ at every point but no Lipschitz condition and not even continuous.
The result that Royden was going for was a theorem of Dini's from 1878.

U. Dini, Fondamenti per la teoretica delle funzione di variabli
  reali, Pisa 1878.

But Dini assumed that $f$ is continuous!
There is a full account in S. Saks, Theory of the Integral (1937) p.~204.
P.S. Royden's first edition was from 1963.  The third edition, which is the only one I have, is from 1988 and the mistake survives in that edition.  Does anyone have a later edition where the problem was corrected?
A: Since $D^+f(x)$ is bounded, let 
$$
D^{+}f(x) = \limsup_{h\to\ 0^+} \frac{f(x+h) - f(x)}{h}=\inf\sup_{h>0}\frac{f(x+h) - f(x)}{h}=M
$$
So given any $\epsilon>0$, there is a $h'$ such that for any $h<h'$
$$
M\leqslant\frac{f(x+h) - f(x)}{h}\leqslant\sup_{h>0}\frac{f(x+h) - f(x)}{h}<M+\epsilon
$$
Hence for any $|x-y|<h$, there is
$$
\left|\frac{f(x)-f(y)}{x-y}\right|<M+\epsilon
$$
i.e
$$
|f(x)-f(y)|<(M+\epsilon)|x-y|
$$
So $f$ is Lipschitz.
