Can you please help me with proving Lipschitz condition?

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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  • $\begingroup$ I corrected a few things in the question. Please make sure that I understood your intention correctly. $\endgroup$ – Ludolila Jun 5 '15 at 14:20
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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jun 5 '15 at 14:21
  • $\begingroup$ How did you fix it? I just typed as the syntax it is. But, it did not apply it. $\endgroup$ – YoungCHOI Jun 5 '15 at 14:22
  • $\begingroup$ $\LaTeX$ code must be between dollar signs, as in a $\LaTeX$ document. $\endgroup$ – Clement C. Jun 5 '15 at 14:23
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    $\begingroup$ @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 $\alpha=1$. $\endgroup$ – 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 $\beta>0$.


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