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The problem:

Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t) - f(x)| \leq |t - x|^2$ for all $t, x$. Prove $f$ is constant.

I believe I have some intuition about why this is the case; i.e. if $t$ and $x$ are very close, ($|t - x| = \epsilon$), then $\sqrt{\epsilon}$ will converge to 0, so $|f(t) - f(x)|$ will also converge to zero if you keep taking $t$ and $x$ closer and closer to each other. However, how do I formalize this argument? Thanks.

Git Gud
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r123454321
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3 Answers3

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For all $x,t\in \Bbb R$ such that $x\neq t$, the equivalence below holds:

$$|f(t) - f(x)| \leq |t - x|^2\iff \left\vert \dfrac{f(t)-f(x)}{t-x}\right\vert\leq |t-x|,$$

taking the limit as $t$ approaches $x$ yields $$\lim \limits_{t\to x}\left(\left\vert \dfrac{f(t)-f(x)}{t-x}\right\vert\right)\leq \lim \limits_{t\to x}|t-x|=0.$$

This proves that $f$ is $\bbox[5px,border:2px solid #FFFFFF]{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$ and $\forall x\in \Bbb R(f'(x)=\bbox[5px,border:2px solid #FFFFFF]{\_})$, thus $f$ is $\bbox[5px,border:2px solid #FFFFFF]{\_\_\_\_\_\_\_\_\_\_\_}$.

Git Gud
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  • Thanks! Does this show that $f'(x) \leq 0$, or that $|f'(x)| \leq 0$? I see in the latter case, we can then conclude that $f'(x) = 0$ and thus the function is constant, but how to get rid of the $f'(x) < 0$ case in the former? Thanks again. – r123454321 Oct 23 '13 at 18:29
  • @RyanYu Since the absolute value function is continuous, $\lim \limits_{t\to x}\left(\left\vert \dfrac{f(t)-f(x)}{t-x}\right\vert\right)=\left\vert \lim \limits_{t\to x}\left(\dfrac{f(t)-f(x)}{t-x}\right)\right\vert$. Does this help? – Git Gud Oct 23 '13 at 18:32
  • Excellent, thanks, makes sense! – r123454321 Oct 23 '13 at 18:38
  • @RyanYu Please note that this is pretty much the same answer Sade gave. – Git Gud Oct 23 '13 at 18:38
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From the given expression We Can Write $\lim_{t\to x}-|t-x|\le \lim_{t\to x}{f(t) - f(x)\over t-x} \leq \lim_{t\to x} |t - x|$, now can you conclude $f'(x)=0\forall x$?

Myshkin
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Alternative (but of course similar) idea:

The assumption is that $g(a,b):=\left|\frac{f(a)-f(b)}{(a-b)^2}\right|$, $a\ne b$, is bounded, indeed by $1$. Now show that, given $a$, $b$, $a\ne0$ and setting $c:=\frac{a+b}2$, you have $g(a,c)\ge2g(a,b)$ or $g(b,c)\ge2g(a,b)$. Conclude that if there are $a$, $b$ with $f(a)\ne f(b)$, then $g$ is unbounded.

Sorry, no derivatives ;)

Carsten S
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