## Winter 1993 #2

Problem Statement: Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfies $|f(x)-f(y)|\leq |x-y|^p$ for every $x,y\in\mathbb{R}$ and for $p>1$ a constant. Prove $f$ is a constant function.

Proof: Let $x\in\mathbb{R}$. Then the absolute value of the derivative is defined to be $|f'(x)|=|\lim\limits_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}|$. By our assumption it follows that

$|f'(x)|=\lim\limits_{h\rightarrow 0}\dfrac{|f(x+h)-f(x)|}{|h|}$

$\leq \lim\limits_{h\rightarrow 0}\dfrac{|(x+h)-x|^p}{|h|}$

$=\lim\limits_{h\rightarrow 0}\dfrac{|h|^p}{|h|}$

Since $p>1$ it follows that $p-1>0$, thus:

$=\lim\limits_{h\rightarrow 0} |h|^{p-1}=0$

Since this limit exists for every $x\in\mathbb{R}$ we may conclude that $f$ is differentiable on $\mathbb{R}$. Furthermore, we may conclude that $0\leq |f'(x)|\leq 0$ for all $x\in\mathbb{R}$ which implies that $|f'(x)|=0$ for all $x\in\mathbb{R}$, thus, $f(x)=0$ for all $x\in\mathbb{R}$. Therefore, $f$ is a constant function on $\mathbb{R}$.

$\Box$

Reflection: This is one of those proofs that really falls straight from the definitions and the hypotheses. If we didn’t have that $p>1$ then we couldn’t conclude that we didn’t have a zero in the denominator. This one was very straightforward and clean.