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Let $f$ be defined for all real $x$ and suppose that $|f(x) - f(y)| \leq |x - y|^{1+ \alpha}$ for all real $x$ and $y$, where $ \alpha > 0$. Prove that $f$ is constant.

Proof: I shall show that the derivative of $f$ is zero. We have that $0 \leq | {f(y) - f(x)} | \leq |x-y|^{1+ \alpha}$. Dividing through by $|x-y|$ we have $0 \leq |\frac{f(y) - f(x)}{y - x}| \leq |y - x|^{\alpha}$ and letting $y \rightarrow x$, we have shown that $f'(x) = \lim_{x \to y}|\frac{f(y) - f(x)}{y - x}| = 0$. Therefore $f$ is a constant.

Jack
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  • You've wedged the limit between 0 and 0, so you can conclude it's zero. The derivative doesn't require the absolute value. You have $0 \leq f'(x) \leq 0$. – Anthony Peter Apr 07 '15 at 04:22
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    If the absolute value of something has limit $0$, then the something has limit $0$. – André Nicolas Apr 07 '15 at 04:22
  • Thanks I wasn't sure if that was true or not! – Jack Apr 07 '15 at 04:23
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    The way to see this is that $x \mapsto |x|$ is a continuous function, so you can take the limit inside. Then you have $|x| = 0$ iff $x=0$. – G. H. Faust Apr 07 '15 at 04:26
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    Suppose it is the year 1517 so you have no calculus. Let $y-x=L $. And for $n\in \mathbb N$ let $x_{j,n}=x+jL/n$ for $0\leq j\leq n.$..... Then $|f(x)-f(y)|=$ $|\sum_{j=0}^{n-1} f(x_{j,n})-f(x_{j+1,n})|\leq$ $ \sum_{j=1}^{n-1}|f(x_{j,n})-f(x_{j+1,n})|\leq $ $n |L/n|^{1+\alpha}=$ $L^{1+\alpha}/n^{\alpha}$ which is arbitrarily close to $0$ if $n$ is large enough. – DanielWainfleet Jul 08 '17 at 01:08

2 Answers2

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Your proof looks fine, but note that $\lim\limits_{x \to y} \left| \frac{f(y)-f(x)}{y-x} \right|$ is equal to $|f'(y)|$, not $f'(x)$.

Newb
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Note that at the end of the proof you write

$f'(x) = \lim\limits_{x\to y} \bigg| \dfrac{f(y) - f(x) }{y - x}\bigg|$ ,

which is not exactly true. The derivative of $f(x)$ can be expressed as

$$ f'(x) = \lim\limits_{x\to y} \dfrac{f(y) - f(x) }{y - x}, $$

thus we have

\begin{equation} \lim\limits_{x\to y} \bigg| \dfrac{f(y) - f(x) }{y - x} \bigg| = \big |\;f'(x)\big| = 0 \end{equation} for any real $x$, which implies that $f'(x) = 0$ everywhere, so that $f$ is a constant.

Vlad
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