# $f$ defined $|f(x) - f(y)| \leq |x - y|^{1+ \alpha}$ Prove that $f$ is a constant. [duplicate]

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.

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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 7 '15 at 4:22
• If the absolute value of something has limit $0$, then the something has limit $0$. – André Nicolas Apr 7 '15 at 4:22
• Thanks I wasn't sure if that was true or not! – Jack Apr 7 '15 at 4:23
• 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 7 '15 at 4:26
• 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 8 '17 at 1:08

## 2 Answers

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 $f'(x)$.

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

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