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Please I need help for this exercise : Let $f : \mathbb R → \mathbb R$, and suppose that $|f(x)−f(y)| \leq (x−y)^2$ for all $x, y$ ∈ $\mathbb R$. Prove that $f$ is constant. First if $a<b$ does $lim a < lim b$? If that is correct I can prove that lim of$ |f(x)−f(y)|$ as $x \rightarrow y$ is $0$ by sandwich theorem and hence $ f$ is constant. But this is a chapter about differentiability I think I should proof that $f'(0)=0$ but I don't know how. What do you think?

User99680
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Guest101
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  • "first if $a<b$ does $\lim a<\lim b$" Not in general. For example, $x^4<x^2$ if $|x|<1$. But $\lim_{x\to0}x^4=0=\lim_{x\to 0}x^2$. What you can say for sure is that if $a<b$ and both $\lim a$, $\lim b$ exist, then $\lim a\le \lim b$, but (as the example shows) equality is possible. – Andrés E. Caicedo Nov 11 '13 at 00:09
  • Because $|f(x)-f(y)|\leq (x-y)^{2}$ with the epsilon-delta defintion we have easily that $f$ is continuous (by finding a $δ<1$).Suppose that $f$ is not constant,then it's image is an interval(due to the fact that $f(\Bbb R)$ is connected),then apply the intermediate value theorem for contradiction. – Haha Nov 11 '13 at 00:43

2 Answers2

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Let's prove it's differentiable first for every $x \in R$. In other words, we want to prove that $$\lim_{y\rightarrow x} \frac{f(y)-f(x)}{y-x}$$ exists. Well consider that $$0 \leq \lim_{y\rightarrow x} \frac{|f(y)-f(x)|}{|y-x|} \leq \lim_{y\rightarrow x} \frac{|y-x|^2}{|y-x|} = \lim_{y\rightarrow x}|y-x| = 0$$ so in particular we've actually shown that $f'(x) = 0$ for any $x\in R$ (why?).

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You have $\lim_{x \to x_0} |\frac{f(x)-f(x_0)}{x-x_0}| \leq |x-x_0|$ so the conclusion follows.

A more difficult question is

Let $f: \mathbb{Q}\rightarrow \mathbb{R}$ be a function such that $$|f(x)-f(y)|\leq (x-y)^{2}$$ for all $x,y \in\mathbb{Q}$. Prove that $f$ is constant.

(for more details on the second problem take a look at: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=813852&sid=4e6ad649f1e45c5dc48362db41b442d4#p813852)

Beni Bogosel
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  • your more difficult question is missing something (the statement of the problem in your link is too). The Dirichlet function ( http://mathworld.wolfram.com/DirichletFunction.html ) satisfies your condition. – Stefan Smith Nov 11 '13 at 01:30