I know this question has been answered before but the way that I am trying to show is different and uses the open balls for the proof. I need a clarification for some part of my proof.
Let $A \subset \mathbb{R}^n$ where $A$ is open and let $f: A \rightarrow \mathbb{R}^n$ be locally Lipschitz on $E$. Prove that $f$ is continuous on $E$.
Definition(locally Lipschitz on $A$): $f$ is locally Lipschitz on $A \subset \mathbb{R}^n$ if and only if $\forall x_0 \in A$ there exist $\delta >0$ such that $B_{\delta}(x_0) \subset A$ and f is Lipschitz on $B_{\delta}(x_0)$.
Now we need definition of Lipschitz on $B_{\delta}(x_0)$ as follows
Definition(Lipschitz on $B_{\delta}(x_0)$):
$f$ is Lipschitz on $A \subset \mathbb{R}^n$ if and only if there exist k>0 such that $\forall x,y \in A$
$$
\|f(x)-f(y)\|\leq k\|x-y\|
$$
Following the definitions makes it super easy to come up with:
$f$ is locally Lipschitz on $A$ then $\forall x_0 \in A$ there exist $\delta >0$ such that $B_{\delta}(x_0) \subset A$ and f is Lipschitz on $B_{\delta}(x_0)$. Therefore there exist a $k$ such that
$$
\|f(x)-f(x_0)\|\leq k\|x-x_0\| \,\,\,\,\, \forall x \in B_{\delta}(x_0)
$$
Then, since $\|x-x_0\| < \delta$, $\|f(x)-f(x_0)\|\leq k\delta=\epsilon \,\,\,\,\, \forall x \in B_{\delta}(x_0)$. $Done!$
But I do not understand two things:
1- Why at this stage we can conclude that $f$ is continuous on $A$? In other words why we can go backward and say $\forall \epsilon>0$ there exist $\delta' > 0 $ such that if $B_{\delta'}(x_0)$ then $B_{\epsilon}(f(x_0))$.
2- Why when we know $f$ is Lipschitz on $B_{\delta}(x_0)$ we do not use $x,y \in B_{\delta}(x_0)$ and we use $x_0$ as one of our point? In other words, can we prove this using $x,y \in B_{\delta}(x_0)$?
Best Answer
Continuity is a local property on $A$. For $f$ to be continuous on $A$, we must show that $f$ is continuous at every fixed point $x_0 \in A$. Moreover, the continuity of $f$ at $x_0$ depends only on the behaviour of $f$ near $x_0$. Put otherwise, when discussing the continuity of $f$ at a fixed point $x_0$, we only care about how $f$ behaves in a neighbourhood of $x_0$. Therefore, $f$ being continuous on a ball $B(x_0,\delta) \subset A$ immediately gives the continuity of $f$ at the point $x_0$.
As for your second question, say we fix $x_0 \in A$ and choose a small ball $B(x_0,\delta) \subset A$ on which $f$ is Lipschitz. Because $f$ is only assumed to be locally Lipschitz, we cannot claim that the inequality $$ ||f(x)-f(y)|| \leq k ||x-y|| $$ holds for all $x,y \in A$. However, we can certainly claim that it holds for every $x,y \in B(x_0,\delta)$. This implies that $f$ is uniformly continuous on the small ball $B(x_0,\delta) \subset A$, which is stronger than simply being continuous there.
Edit:. Fix $x_0 \in A$ and let $B(x_0,\delta) \subset A$ be an open ball on which $f$ is Lipschitz. Then, we can find $L > 0$ such that $$ ||f(x)-f(y)||\leq L||x-y||, \quad \forall x,y \in B(x_0, \delta). $$ In particular, because $x_0 \in B(x_0,\delta)$, we have $$ ||f(x)-f(x_0)||\leq L||x-x_0||, \quad \forall x\in B(x_0, \delta). $$ Using this, we will show that $f$ is continuous at $x_0$. Let $\epsilon > 0$ and choose $\delta^\prime > 0$ such that $$ 0 < \delta^\prime < \frac{\epsilon}{L} \quad \text{and} \quad \delta^\prime < \delta. $$ Then, if $|x-x_0| < \delta^\prime$ one also has $x \in B(x_0,\delta)$ so that \begin{align*} ||f(x)-f(x_0)|| \leq L||x-x_0|| < L\delta^\prime < \epsilon. \end{align*} Or, rather, $f$ sends $B(x_0,\delta^\prime)$ into the ball $B(f(x_0),\epsilon)$. This means that $f$ is continuous at $x_0$. By our remarks above the edit, $f$ is also continuous on $A$.