Hölder continuity of $\frac1x$ I have a question. Is the function $f(x)=1/x$ Hölder continuous if $x\in (\varepsilon,+\infty),\ \varepsilon>0$?
 A: For a fixed value of $\epsilon$, the derivative of $\frac1x$ is bounded on $(\epsilon, \infty)$, so the function is actually Lipschitz continuous.
A: Since $f$ is differentiable in $(\varepsilon,+\infty)$, you can apply mean value theorem:
$$
|1/x-1/y| = |-1/z^2|\cdot|x-y|\leq 1/\varepsilon^2 |x-y|,
$$
where $z\in [x,y]$.
A: Let $x,y\in (\epsilon,\infty)$ for some $\epsilon>0$.
Observe that $|f(x)-f(y)|=\left|\dfrac{1}{x}-\dfrac{1}{y}\right|=\left|\dfrac{x-y}{xy}\right|\le \dfrac{1}{\min \{x,y\}}|x-y|\le \dfrac{1}{\epsilon^2}|x-y| $.
Therefore for each $x,y\in (\epsilon,\infty)$, there exists $\delta \left( =\dfrac{1}{\epsilon}\right) >0$ and $\alpha (=1)>0$ such that $|f(x)-f(y)|\le \delta |x-y|^{\alpha}$. Hence $f$ is Hölder continuous.
A: Recall that $f$ is $\alpha$-Holder continuous if, only if, 
$$
\sup_{x,y\in (\epsilon, \infty)}\frac{| f(x)-f(y)|}{|x-y|^\alpha}<\infty
$$
Note that
$$
\sup_{x,y\in (\epsilon, \infty)}\frac{\left|\frac{1}{x}-\frac{1}{y}\right|}{|x-y|^\alpha}
=
\sup_{x,y\in (\epsilon, \infty)}\frac{\frac{|x-y|}{|xy|}}{|x-y|^\alpha}
=
\sup_{x,y\in (\epsilon, \infty)}\frac{|x-y|^{1-\alpha}}{|x  \cdot y|}=+\infty
\quad \mbox{ if }\; \alpha < 1
$$
If $\alpha=1$, then 
$$
\sup_{x,y\in (\epsilon, \infty)}\frac{\left|\frac{1}{x}-\frac{1}{y}\right|}{|x-y|}
=
\sup_{x,y\in (\epsilon, \infty)}\frac{1}{|x  \cdot y|}=\frac{1}{\epsilon^2}
$$
