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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ some function that all $x$ and $y$ in $\mathbb{R}$ satisfies: $$\left|f(x)-f(y)\right| \le (x-y)^2$$

  • Prove that f is differentiable in any point in $\mathbb{R}$.
  • Prove that f is constant.
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    $\begingroup$ FYI, it's "differentiable" :) $\endgroup$
    – Jim
    Feb 4, 2013 at 16:47
  • $\begingroup$ @Jim thank's :) $\endgroup$
    – Nicole
    Feb 4, 2013 at 16:55
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    $\begingroup$ A function with the property that $|f(x)-f(y)| \le C |x-y|$ for some $C \gt 0$ and all $x,y$ is said Lipschitz continuous. There's a story that makes the rounds in math grad schools that a PhD student was defending a thesis on marvelous properties of functions with Lipschitz exponent greater than 1, meaning something of the sort illustrated here by a square power. Possibly apocryphal, as the story goes an implication was pointed out by a defense committe member other than the thesis adviser, that all such functions are (as here) constant. $\endgroup$
    – hardmath
    Feb 4, 2013 at 17:01
  • $\begingroup$ For another approach, see Davide Giraudo's answer in this post. $\endgroup$ Feb 4, 2013 at 17:04

5 Answers 5

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You have $|f(x)-f(y)|\le |x-y||x-y|$ then $\displaystyle\frac{|f(x)-f(y)|}{|x-y|}\le|x-y|$ Just take the limit as $x\to y$ , you get $|f'(y)|\le 0$(then $f'(y)=0)$ for all $y$ then $f$ is constant

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  • $\begingroup$ Please edit to use the absolute value in the denoninator $\endgroup$ Feb 4, 2013 at 16:56
  • $\begingroup$ @Barbara Thank you for pointing it out $\endgroup$
    – user10444
    Feb 4, 2013 at 16:59
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I would like to point out that it is possible to prove the assertion without proving differentiability in the first place.

For each x we have

$$|f(x)-f(0)| = \left|\sum_{k=1}^n f\left(\frac{k}{n} \cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$

Thus by the triangle inequality

$$|f(x)-f(0)| \le \sum_{k=1}^n \left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$

Since for each $k$

$$\left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right| \le \left|\frac{k}{n}\cdot x-\frac{k-1}{n}\cdot x\right|^2 = \left|\frac{x}{n}\right|^2$$

It follows that

$$|f(x)-f(0)| \le \sum_{k=1}^n \frac{x^2}{n^2} = \frac{x^2}{n}$$

And thus

$$|f(x)-f(0)|\le \limsup_{n\rightarrow\infty} \frac{x^2}{n} = 0$$

Hence for all $x$

$$f(x)=f(0)$$

Thus f is constant. It follows that f is differentiable everywhere ;)

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By definition of what it means to be differentiable, you want to prove that the limit $$ \lim_{x\to y} \frac{f(x)- f(y)}{x-y} $$ exists for all $y\in \mathbb{R}$. That will follow (in this case) from showing that $$ \lim_{x\to y} \left\lvert\frac{f(x)- f(y)}{x-y}\right\rvert $$ exists. Now you have that $$ 0\leq \left\lvert\frac{f(x)- f(y)}{x-y}\right\rvert = \frac{\lvert f(x)- f(y)\rvert}{\lvert x-y\lvert} \leq \frac{\lvert x - y\rvert^2}{\lvert x - y\rvert} = \lvert x - y\rvert. $$ Now you have $\lvert x - y\rvert \to 0$ as $x \to y$. So by the Squeeze Theorem you must also have $$ \lim_{x\to y} \left\lvert\frac{f(x)- f(y)}{x-y}\right\rvert = 0 $$ And so $$ \lim_{x\to y} \frac{f(x)- f(y)}{x-y} = 0 $$ That means that $f$ is differentiable and that the derivative at any number $y$ is zero: $f'(y) = 0$.

As others have already mentioned this means that $f$ must be a constant: If you had $f(x) \neq f(y)$ for some $x$ and $y$. Then by the Mean Value Theorem you would have a $c$ between $x$ and $y$ such that $0\neq f(x) - f(y) = f'(c)(x-y) = 0$. This is a contradiction, so indeed $f(x) = f(y)$ for all $x$ and $y$.

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  • $\begingroup$ It's not the same to show the absolute value of the difference quotient has a limit, but to show the difference quotient tends to zero it suffices to show the absolute value of the difference quotient tends to zero. $\endgroup$ Feb 4, 2013 at 19:29
  • $\begingroup$ @MatthewLeingang: I edited. Thanks. $\endgroup$
    – Thomas
    Feb 4, 2013 at 19:37
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For each $x\in\mathbb R$, for each $y\in\mathbb R$ with $y\neq x$, we have $0\leq |\frac{f(x)-f(y)}{x-y}|\leq |x-y|$. Let $y\to x$. Then by squeezing we obtain $f'(x)=0$ for all $x\in\mathbb R$ so that $f$ is constant.

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  • $\begingroup$ I think you mean $y \to x$. $\endgroup$
    – Jim
    Feb 4, 2013 at 16:51
  • $\begingroup$ Shouldn't there be absolute values as in $\le \vert x-y \vert$? $\endgroup$ Feb 4, 2013 at 16:54
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Let $a\in\Bbb R$ be fixed and $x\neq a$ then $$\left| \frac{f(x)-f(a)}{x-a}\right|\le|x-a|$$ Since $\lim_{x\to a}|x-a|=0$, we get $\lim_{x\to a} \left| \frac{f(x)-f(a)}{x-a}\right|=0$. So $f$ is differentiable at $a$.

And $f'(a)=0$ for all $a$, By mean value theorem we get $$f(x)-f(y)=f'(c)(x-y)$$ for some $c$. Since $f'(c)=0$, we get $f(x)=f(y)$, So $f$ is constant.

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