If $f:\mathbb R\rightarrow \mathbb R$ is such that $|f(x)-f(y)|\leq|x-y|^\sqrt2$ $\forall x$and $y\in R$

Question

If $f:\mathbb R\rightarrow \mathbb R$ is such that $$|f(x)-f(y)|\leq|x-y|^\sqrt2$$ $\forall x$ and $y\in \mathbb R$,then $f$ is constant.

MY ATTEMPT:$f : [a,b] \rightarrow \mathbb{R}$ such that $$|f(x)-f(y)|\leq c|x-y|^k$$ where $c\geq0$ and $k>1$ ,then $f$ is constant.My question is that "is it true for my above question and how can we prove it?Thank you.

Answer 1

Consider the limit $$\lim_{y\to x}\frac{f(y)-f(x)}{y-x}.$$ Clearly, this limit exists and is equal to zero. On the other hand, this limit is the definition of $f'(x)$, hence $\forall x\in \Bbb R \, f'(x)=0$. What can you say about a function whose derivative is identically zero?

Answer 2

Hint: $$\lim\limits_{h\rightarrow 0}\left\lvert\dfrac{f(x+h)-f(x)}{h}\right\rvert\leq\lim\limits_{h\rightarrow 0}\enspace\lvert h^{k-1}\rvert=0,$$ as $k>1$.

Answer 3

$\lim\limits_{x\rightarrow y}\left\lvert\dfrac{f(x)-f(y)}{x-y}\right\rvert\leq\lim\limits_{x\rightarrow y}\enspace\lvert (x-y)^{\sqrt2-1}\rvert=0,$ as $\sqrt2-1>0$ i.e $f'(x)=0 \forall x\in \mathbb R$. So$f$ is constant.