If $f(x)$ satisfies a Lipschitz condition of order $\alpha>1$ on $[a, b]$, prove that $f(x)$ is a constant function on $[a, b]$.

Question

If $f(x)$ satisfies a Lipschitz condition of order $m>1$ on $[a, b]$, prove that $f(x)$ is a constant function on $[a, b]$.

Lipschitz condition of order $m>1$ is $$|f(x)-f(y)|<\lambda |x-y|^m~ ~~~~\forall x,y\in [a, b]$$ where $\lambda>0$. But how to show that $f'(x)=0$?

Answer 1

$\lvert \dfrac{f(a+h)-f(a)}{h}\rvert<\lambda \dfrac{h^m}{h}=\lambda h^{m-1}\to 0$ as $h\to 0$ as $m-1>0$

Since $a$ is arbitrary $\implies f^{'}(a)=0\forall a\implies f$ is constant