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Let $f : R \rightarrow R $
$\lvert f(x)-f(y) \rvert \le (x-y)^2, \forall x,y \in R $

Any sort of help is appreciated! I know I am not suppose to ask for the entire solution, so I will ask for strong hints.

Trying2LearnMath
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  • This question is just a special case of http://math.stackexchange.com/questions/124263/a-condition-for-a-function-to-be-constant – Silverfish Apr 28 '15 at 09:39

3 Answers3

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HINT: For all $n\in\Bbb Z^+$,

$$\begin{align*} |f(x)-f(x+a)|&\le\sum_{k=0}^{n-1}\left|f\left(x+\frac{ka}n\right)-f\left(x+\frac{(k+1)a}n\right)\right|\\ &\le\sum_{k=0}^{n-1}\left(\frac{a}n\right)^2\\ &=\frac{a^2}n\;. \end{align*}$$

Brian M. Scott
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    +1 for no derivative. However I'd rather say the same with more words and less formaulae: explicitly describe dividing the $[x,y]$ range of length $L=|y-x|$ into $n$ parts of equal lengths $d=\frac Ln$, then the sum of $f$ increments over all parts is directly $|f(x)-f(y)| \le nd^2=\frac {L^2}n$ which tends to $0$ for growing $n$. – CiaPan Apr 28 '15 at 06:51
  • @CiaPan: The whole point was not to say it explicitly, but rather to give, as the OP said, a strong hint -- something that would point the way reasonably clearly but still require some thought. – Brian M. Scott Apr 28 '15 at 06:54
  • Well, seems I haven't read the whole question. :( – CiaPan Apr 28 '15 at 07:30
  • In my opinion, more words and less formulae would be a better hint than this one. This is almost an answer which can be copied directly without a thought, but the same with words would help the OP understand the point. – JiK Apr 28 '15 at 15:22
  • @JiK: It requires significant addition in the way of explanation in order to become an answer that I would accept for full credit. And in order to produce that explanation, the reader has to understand the point of the calculation; until that’s understood, it’s just a string of symbols. My goal is to give the reader enough help to arrive at that understanding on his own without explaining it. – Brian M. Scott Apr 28 '15 at 17:41
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$$ \bigg|\frac{f(x)-f(y) }{x-y} -0 \bigg|\leq |x-y|$$

So if $x$ goes to $y$ then $f'(y)=0$ for all $y$

HK Lee
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    Fair enough; I was still reacting to the other answer, I guess, and didn’t think through what you were actually suggesting. – Brian M. Scott Apr 28 '15 at 02:53
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Think about a few things:

  1. The limit definition of the derivative,
  2. The derivative of a constant function

And let $x$ tend to $0$