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Proof of a simple property of real, constant functions.

Suppose $|f(x)-f(y)|\leq (x-y)^2$ for all $x,y\in\mathbb{R}$. Show f is differentiable.

This follows intuitively, the derivative $2(x-y)$ is defined on $\mathbb{R}$. How do I show this formally?

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marked as duplicate by t.b., Did, Asaf Karagila, Carl Mummert, Zev Chonoles Apr 12 '12 at 22:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

"the derivative $2(x-y)$ is defined on $\mathbb{R}$" does not make any sense. – Martin Brandenburg Apr 12 '12 at 18:46
Hint: Let $x=a+h$ and $y=a$. Then $|f(a+h)-f(a)| \le h^2$. – André Nicolas Apr 12 '12 at 18:48
This problem is easier than this one:… . – Beni Bogosel Apr 12 '12 at 18:57
I restored the content of the question because it has two answers already, including one that is quite detailed. – Carl Mummert Apr 12 '12 at 19:40

You have to show that, for every $x \in \mathbb{R}$, the limit of $\frac{f(x+h)-f(x)}{h}$ as $h \to 0$ exists. In fact, the limit is zero. In order to show this, just begin with $\left|\frac{f(x+h)-f(x)}{h}\right| \leq \dotsc$.

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You can write this inequality as: $$-(x-y)^2\leq f(x)-f(y)\leq (x-y)^2$$ Assume that $x>y$. Then $$-(x-y)\leq \frac{f(x)-f(y)}{x-y}\leq (x-y)$$Taking the limits as $x$ approaches $y$ and respecting the case $y>x$ you come to the conclusion that $f'(y)=0$ for all $y$. So, $f$ is differentiable.

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Then $f$ is constant (in particular, $f$ is differentiable).

To show this, consider $x\ne y$, split the interval between $x$ and $y$ into $n\geqslant1$ parts of length $|x-y|$ and apply the triangular inequality. This yields $$ |f(x)-f(y)|\leqslant\sum_{k=1}^n\left|f\left(x+\frac{k-1}n(y-x)\right)-f\left(x+\frac{k}n(y-x)\right)\right|. $$ Now, apply the hypothesis to each of these intervals. The result is $$ |f(x)-f(y)|\leqslant\sum_{k=1}^n\frac{|y-x|^2}{n^2}=\frac1n|y-x|^2. $$ When $n\to\infty$, this proves that $f(x)=f(y)$.

Note that the same result holds as soon as $|f(x)-f(y)|\leqslant C|x-y|^a$ for every $x$ and $y$, for some $a\gt1$.

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+1 for pointing it out. However, this seems simpler to me: Divide through by $(x-y)$, let $x \to y$ and conclude that $f' = 0$. – t.b. Apr 12 '12 at 19:54
@t.b. One could argue that it is easier to avoid differentiation altogether. – Did Apr 12 '12 at 19:56
Indeed, one could... :) – t.b. Apr 12 '12 at 20:00
@t.b. Did you modify your first comment? I do not see the easier that was in it... :-) – Did Apr 12 '12 at 20:12
Yes, I did, sorry about that; I submitted the edit more or less at the same time as your comment appeared and I thought the change wasn't important enough to try and reconstruct what exactly I wrote before. – t.b. Apr 12 '12 at 20:24

Have a look at this. Read the description and the top response. That may give you a clue or two.

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