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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 '12 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 '12 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 '12 at 6:32
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    $\begingroup$ Somewhat relevant to the above comments: meta.math.stackexchange.com/questions/1878/… $\endgroup$ – Jonas Meyer Jun 30 '12 at 6:37
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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 '12 at 6:37
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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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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? $\endgroup$ – creepyrodent Nov 7 '18 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 '18 at 17:24
  • $\begingroup$ Oh, right. The order things appeared confused me. Thanks. $\endgroup$ – creepyrodent Nov 7 '18 at 17:36
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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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