Suppose that $│f(x) - f(y)│≤│x-y│^2$ for all $x,y \in \mathbb{R}$
Proof that $f$ is constant by computing the derivative $f$
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This just says that the derivative is bounded by 2. – copper.hat Nov 24 '13 at 08:14
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@copper.hat : I think that is meant to be $|x-y|^2$ – Prahlad Vaidyanathan Nov 24 '13 at 08:15
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1Shouldn't it be given that the function is differentiable? – Priyatham Nov 24 '13 at 08:16
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@PrahladVaidyanathan: I think you are right. – copper.hat Nov 24 '13 at 08:25
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Duplicate 1 and Duplicate 2. – Git Gud Nov 24 '13 at 08:50
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$$ \lim_{h\to 0} \left | \frac{f(x+h) - f(x)}{h}\right | \leq \lim_{h\to 0} \frac{|h|^2}{|h|} = 0 $$ Hence, $f' \equiv 0$
Prahlad Vaidyanathan
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Hint. If $|f(x) - f(y)| \leq |x - y|^2$ for all real $x$ and $y$, then we have $-|x - y|^2 \leq f(x) - f(y)$ and $f(x) - f(y) \leq |x - y|^2$ for all real $x$ and $y$. Now the squeeze theorem states, roughly, that if $f(x) \leq g(x)$ and $g(x) \leq h(x)$ for all $x$ in some interval containing $a$, then $\lim_{x \rightarrow a} g(x) = \ell$ whenever $\lim_{x \rightarrow a}f(x) = \ell$ and $\lim_{x \rightarrow a}h(x) = \ell$. From the given inequalities, we obtain $$ \frac{-|x - y|^2}{x - y} \leq \frac{f(x) - f(y)}{x - y} \leq \frac{|x - y|^2}{x - y} \text{ for all } x \neq y $$ Now recall the definition of a derivative and consider the cases $x \geq a$ an $a > x$ to find the required limits. You should find that the middle limit is squeezed between two functions having $0$ as their limits as $x$ approaches $y$. These limits give the derivatives of the functions at the arbitrary point $y$.
emi
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