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Suppose $f$ is defined on all of ${\Bbb R}$, and satisfies, $|f(x)-f(y)|\leq(x-y)^2$ for all $x,y\in {\Bbb R}$. Prove that $f$ is constant.

Basically I have,

$|f(x)-f(y)|\leq(x-y)^2 \Rightarrow \frac{|f(x)-f(y)|}{(x-y)}\leq(x-y)$.

$\Rightarrow \lim_{\ x\to\ y}\frac{|f(x)-f(y)|}{(x-y)}\leq \lim_{\ x\to\ y}(x-y)$.

$\Rightarrow f^{'}(x)\leq 0, \ \forall x\in{\Bbb R}$.

Case 1: if $f^{'}(x)= 0$ then we're done.

Case 2: if $f^{'}(x)< 0$???

Well I don't even know if I'm going about this right. Any hints?

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1 Answer 1

up vote 1 down vote accepted

You've really shown more: Using the fact that $(x - y)^2 = |x - y|^2$, we have

$$0 \le \left|\frac{f(x) - f(y)}{x - y}\right| \le |x - y|$$

Now take the limit as $y \to x$, and we see from the Squeeze (Sandwich) Theorem that

$$0 \le |f'(x)| \le 0$$

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AHHh, I was almost there. Thanks a lot!! –  Rod Nov 10 '13 at 6:25
    
@Faraad You're very welcome. –  user61527 Nov 10 '13 at 6:27

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