# Hard problem about Lipschitz condition. This is a very unusual case. [closed]

Let: $f\colon\mathbb{R}\longrightarrow \mathbb{R}$ satisfy $\lvert f(x)-f(y) \rvert\leq C\lvert x-y\rvert^\alpha$ for some constants $C\gt 0$, $0\lt\alpha\lt\infty$, and for any $x,y\in\mathbb{R}$.

(1) Prove that if $0\lt\alpha\le1$, then $f$ is uniformly continuous.

(2) Prove that if $1\lt\alpha\lt\infty$, then $f$ is a constant.

## closed as off-topic by Andrés E. Caicedo, Clement C., Jack D'Aurizio, Siminore, user147263 Jun 5 '15 at 17:42

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Clement C., Jack D'Aurizio, Siminore, Community
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• I corrected a few things in the question. Please make sure that I understood your intention correctly. – Ludolila Jun 5 '15 at 14:20
• Please improve formatting by using $\LaTeX$ and show you efforts. It would be a good idea to remove the begging part from the title, too. – Jack D'Aurizio Jun 5 '15 at 14:21
• How did you fix it? I just typed as the syntax it is. But, it did not apply it. – YoungCHOI Jun 5 '15 at 14:22
• $\LaTeX$ code must be between dollar signs, as in a $\LaTeX$ document. – Clement C. Jun 5 '15 at 14:23
• @YoungCHOI By the way, for your cultural growth, that hypothesis goes under the name of Holder condition, of which the Lipschitz condition is the special case $\alpha=1$. – user228113 Jun 5 '15 at 14:50

1. Use the definition of uniform continuity. For any fixed $\varepsilon > 0$, you want to find $\delta=\delta(\varepsilon) > 0$ such that, for any $x,y\in\mathbb{R}$, $\lvert x-y\rvert \leq \delta$ implies $\lvert f(x)-f(y)\rvert \leq \varepsilon$. Using your hypothesis, why is it sufficient to choose $\delta$ such that $C\delta^\alpha \leq \varepsilon$? (And can you conclude from there?)
2. Prove that $f$ is then differentiable everywhere, by fixing an arbitrary $x$, taking $y=x+h$ and letting $h\to 0$. This will in particular show you that $f^\prime(x)$ not only exists, but equals 0. You will need the assumption $\alpha > 1$ for this, since then $\lvert h\rvert^\alpha = \lvert h\rvert^{1+\beta}$ with $\beta>0$.