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Let $\mathbb R$ be the set of real numbers and $f: \mathbb R \rightarrow \mathbb R$ be such that for all $x$ and $y$ in $\mathbb{R}$, $$|f(x)-f(y) |\leq |x-y|^3.$$ Prove that $f(x)$ is a constant.

This is a new type of problem for me and I feel I am missing some trick to simplify the given expression. Any help??

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  • $\begingroup$ The stated condition means that $f$ has a derivative everywhere, which can be found by using the definition of derivative. $\endgroup$
    – Will Jagy
    Commented Aug 17, 2014 at 5:08
  • $\begingroup$ Related questions: link 1, link 2, link 3, link 4, link 5 $\endgroup$
    – Pedro
    Commented Aug 17, 2014 at 5:45
  • $\begingroup$ And also link 6. $\endgroup$
    – WimC
    Commented Aug 17, 2014 at 10:18

2 Answers 2

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The limit definition of the derivative says that $f'(y) = \displaystyle\lim_{x \to y}\dfrac{f(x)-f(y)}{x-y}$ (provided the limit exists).

Here, we have that $\left|\dfrac{f(x)-f(y)}{x-y}\right| < |x-y|^2$. What does this tell you about the derivative of $f$?

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  • $\begingroup$ Now, I understand the value of derivative is zero, but I am still confused about the absolute value. $\endgroup$
    – user34304
    Commented Aug 17, 2014 at 5:13
  • $\begingroup$ What about the absolute value is confusing? $\endgroup$
    – JimmyK4542
    Commented Aug 17, 2014 at 5:23
  • $\begingroup$ @JimmyK4542: check the fundamental definition of the derivative at a point $a$. $\endgroup$
    – Mhenni Benghorbal
    Commented Aug 17, 2014 at 7:43
  • $\begingroup$ Why the down vote? $\endgroup$
    – copper.hat
    Commented Aug 17, 2014 at 17:55
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Here is a version without derivatives.

Choose $x,y$. Let $t_k = x+{k \over n} (y-x)$, for $k=0,...,n$.

Then $|f(x)-f(y)| \le \sum_{k=0}^{n-1} |f(t_{k+1})-f(t_k)| \le (x-y)^3 \sum_{k=0}^{n-1} {1 \over n^3} = {1 \over n^2}(x-y)^3$.

Since this is true for all $n$, we have $|f(x)-f(y)| = 0$, or in other words, $f(x) = f(y)$.

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  • $\begingroup$ Why the down vote? $\endgroup$
    – copper.hat
    Commented Aug 17, 2014 at 17:54
  • $\begingroup$ I was wondering how you get that $(x-y)^3 \sum_{k=0}^{n-1} {1 \over n^3} = {1 \over n^2}(x-y)^3$? $\endgroup$
    – Skm
    Commented Aug 7, 2019 at 19:00
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    $\begingroup$ I'm not sure what you are asking. $\sum_k {1 \over n^3} = n {1 \over n^3} = {1 \over n^2} $. $\endgroup$
    – copper.hat
    Commented Aug 7, 2019 at 19:11
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    $\begingroup$ Sorry, I meant not a function of $k$. $\endgroup$
    – copper.hat
    Commented Aug 7, 2019 at 19:26
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    $\begingroup$ @K.M: Yes indeed. Thanks for catching that! $\endgroup$
    – copper.hat
    Commented Aug 7, 2019 at 23:27

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