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.

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