$A function satifying $|f(x)-f(y)|\leq |x-y|^2$ must be constanct.$

Question

$A function satifying $|f(x)-f(y)|\leq |x-y|^2$ must be constanct.$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfy $$|f(x)-f(y)|\leq |x-y|^2, \forall x,y\in \mathbb{R}.$$

Prove that $f$ is constanct.

Real Analysis Functional Analysis Calculus

Brandon 1212

23

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Let $x\in \mathbb{R}$. Then
\[|f'(x)|=|\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}|=\lim_{\Delta x \rightarrow 0}|\frac{f(x+\Delta x)-f(x)}{\Delta x}|\]
\[\leq \lim_{\Delta x \rightarrow 0}\frac{|x+\Delta x-x|^2}{|\Delta x|}=\lim_{\Delta x \rightarrow 0}\frac{|\Delta x|^2}{|\Delta x|}\]
\[=\lim_{\Delta x \rightarrow 0}|\Delta x|=0.\]
Hence $f'(x)=0$, for all $x\in \mathbb{R}$. Therefore $f$ must be constant.

Erdos

4.8K

$A function satifying $|f(x)-f(y)|\leq |x-y|^2$ must be constanct.$

Answer

Related Questions

Search