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Problem: Let $f$ be defined for all real $x$, and suppose that

$$|f(x)-f(y)|\le (x-y)^2$$

for all real $x$ and $y$. Prove $f$ is constant.

Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 1.

  • 2
    This has been asked before on the site, with roughly the same answers (based on the derivative like Potato's or on the triangular inequality like Frank's). Potato: I wonder if asking questions you answer yourself immediately is part of a Grand Plan of yours.2012-06-30
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    To be honest about my motives: writing up answers is good proofwriting practice, and occasionally someone spots an error I didn't. It's very educational.2012-06-30
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    Is this a declared goal of the site?2012-06-30
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    @did Yes. ${}{}{}{}$2012-06-30
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    @MattN. Would you know any reference for the fact that it is? (*This* in my previous comment refers to *proofwriting practice*.)2012-06-30
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    @did No apparently not. But as I understand it the site is here to help people improve their mathematical skills. And proofwriting is one.2012-06-30
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    Somewhat relevant to the above comments: http://meta.math.stackexchange.com/questions/1878/are-please-check-my-proof-type-of-questions-proper2012-06-30
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    @did: Perhaps someone should start a meta thread? Feel free to name me explicitly and link to my post history. I will stop if people want me to.2012-06-30
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    *ducks away* ${}{}{}$2012-06-30
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    @did I eventually find that his latest posts are probably consecutive exercises on Rudin's book.2012-06-30
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    Potato: As an aside, I note that in this stream of questions you omit any indication about what you tried, what you know, and so on, which is definitely recommended on the site. @FrankScience: Yes, this is mentioned in the questions.2012-06-30
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    @did But I don't think they're all good questions, also answering one's own question is [encouraged](http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/)2012-06-30
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    @Frank: yes, *answering* one's own problem. Thanks for the thread (which does not seem to have precisely the present situation (posting one by one the exercises from a textbook) in mind...).2012-06-30
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    @did We are saying the same thing (PS: I find that *also* in my comment is really *although*).2012-06-30

3 Answers 3

8

It suffices to show that $f'(x)=0$ for all $x\in\mathbb{R}$. We see that the given condition implies

$$\left| \frac{f(x)-f(y)}{x-y} \right| \le |x-y|.$$

So in a $\delta$-neighborhood of $x$, the quotient in definition of the derivative is less than $\delta$. So the limit is 0, and we are done.

19

For any $x\in\mathbb{R}$, $$ \begin{align} |f'(x)| &=\lim_{h\to0}\frac{|f(x+h)-f(x)|}{|h|}\\ &\le\lim_{h\to0}\frac{h^2}{|h|}\\ &=0 \end{align} $$ Therefore, $f$ is constant.

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    But why is $f$ differentiable?2018-11-07
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    Because the limit, which is the definition of the derivative, exists (and equals $0$).2018-11-07
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    Oh, right. The order things appeared confused me. Thanks.2018-11-07
15

Here's a proof more elementary.

Let $c=f(0)$, we have to prove that $f(x)=c$ whenever $x\neq0$. Supposing that $n$ is an arbitrary positive integer, we have $$\left|f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right|\le\left(\frac{m+1}nx-\frac mnx\right)^2=\frac{x^2}{n^2}$$ Hence \begin{align*} |f(x)-f(0)| \;&=\;\left|\,\sum_{m=0}^{n-1}\left(f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right)\,\right|\\ &\le\;\sum_{m=0}^{n-1}\,\left|f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right|\\ &\le\;\frac{x^2}n \end{align*} Let $n\to\infty$, we have $|f(x)-f(0)|=0$, thus $f(x)=c$.