Show that if $f$ is Lipschitz of order $\alpha > 0$ at $x$, then $f$ is continuous at $x$

Question

A function is called Lipschitz of order $\alpha$ at $x$ if there is a constant $C>0$ for which $|f(x)-f(y)| \leq C|x-y|^{\alpha}$ for all $y$ in an open interval containing $x$. Show that if $f$ is Lipschitz of order $\alpha > 0$ at $x$, then $f$ is continuous at $x$.

Since $f$ is Lipschitz of order $\alpha > 0$ at $x$, we have that there exists a constant $C > 0$ for which $|f(x)-f(y)| \leq C|x-y|^{\alpha}$. Now to show that $f$ is continuous at $x$ we have to show that $\displaystyle \lim_{y \to x} f(y) = f(x)$. The Lipschitz form seems to limit how fast the function can grow, so I need to somehow relate that to the definition of the limit but I am unsure how to.

Answer 1

Let $\;\epsilon>0\;$ and choose $\;\delta=\sqrt[\alpha]\frac\epsilon C\;$ ( observe that it is given $\;\alpha>0\;$ ), so

$$|x-y|<\delta\implies |f(x)-f(y)|\stackrel{\text{given}}<C|x-y|^\alpha<C\delta^\alpha=C\frac\epsilon C=\epsilon$$

and this implies $\;f(y)\xrightarrow[y\to x]{}f(x)\;$ and we've finished

Answer 2

By definition, $f$ is continuous at $x_o$ if $\forall ϵ>0$, there exists $\delta>0$ such that $|x- x_o| < \delta \Rightarrow |f(x) - f(x_o)| < \epsilon$

Let $\epsilon >0$ be given and let $y \in B_{\delta}(x) \Rightarrow |x - y| < \delta$

Then $|f(x) - f(y)| \leq C|x- y|^{\alpha} < C\delta^{\alpha}$

Pick $\delta = {(\dfrac{\epsilon}{C})}^{1/\alpha}$

Then it follows $|f(x) - f(y)| < \epsilon$

Lipschitz implies continuity

"Drill baby drill!"