Hölder continuity of $\frac1x$

Question

I have a question. Is the function $f(x)=1/x$ Hölder continuous if $x\in (\varepsilon,+\infty),\ \varepsilon>0$?

Answer 1

For a fixed value of $\epsilon$, the derivative of $\frac1x$ is bounded on $(\epsilon, \infty)$, so the function is actually Lipschitz continuous.

Answer 2

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 $z\in [x,y]$.

Answer 3

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.