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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.

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    $\begingroup$ 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. $\endgroup$
    – Did
    Jun 30, 2012 at 6:26
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    $\begingroup$ 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. $\endgroup$
    – Potato
    Jun 30, 2012 at 6:27
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    $\begingroup$ @MattN. Would you know any reference for the fact that it is? (This in my previous comment refers to proofwriting practice.) $\endgroup$
    – Did
    Jun 30, 2012 at 6:32
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    $\begingroup$ @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. $\endgroup$
    – Potato
    Jun 30, 2012 at 6:37
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    $\begingroup$ @did But I don't think they're all good questions, also answering one's own question is encouraged $\endgroup$
    – Yai0Phah
    Jun 30, 2012 at 6:59

3 Answers 3

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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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    $\begingroup$ But why is $f$ differentiable? $\endgroup$
    – user561334
    Nov 7, 2018 at 17:22
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    $\begingroup$ Because the limit, which is the definition of the derivative, exists (and equals $0$). $\endgroup$
    – robjohn
    Nov 7, 2018 at 17:24
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    $\begingroup$ Oh, right. The order things appeared confused me. Thanks. $\endgroup$
    – user561334
    Nov 7, 2018 at 17:36
  • $\begingroup$ This is a fantastic example of turning what might be a messy proof into something so elegant, succinct, and clear it's irresistible. I assume this was not your first draft, but rather a rework of a rougher proof. Could you add a bit of how you went from a rougher proof to this? E.g. imagine you started with this, how did you get to your proof? Interestingly, the proof begins not from the given property ($\leq (x-y)^2$), nor from the definition of derivative, but in the middle, with $|f'(x)|$. $\endgroup$ Dec 19, 2022 at 14:32
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    $\begingroup$ We want to show that $f’$ is close to $0$. The cleanest way to show $f’$ is close to $0$ is to show $|f’|$ is small (this avoids having to show two inequalities). $\endgroup$
    – robjohn
    Dec 20, 2022 at 5:12
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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 $f(x)=c$.

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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.

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