show that f is constant on I
Let $f$ be defined on the interval $I$. all we know about $f$ is that there is a constant $K\gt 0$ so that $|f(a)-f(b)|\le K|a-b|^2$ for all $a$,$b$ on $I$.
Q: show that $f$ is constant on $I$
hint: start with calculating the derivative using the definition?
how do I go about differentiating this inequality using the definition, and can I use $a=x$ and $b=x+h$ ? if yes then how
Solution 1:
As you have mentioned $$\lim_{h \to 0} |\frac{f(x+h)-f(x)}{h}| \le \lim _{h \to 0} k|h| $$
Conclude from here .