# How to show locally Lipschitz on $A \subset \mathbb{R}^n$ implies continuity on $A$?

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 $$x,y \in B_{\delta}(x_0)$$?

• You are suffering from a severe case of clash of variables. Don't use the same $\delta$ to represent the radius of the ball in the definition of locally Lipschitz and the $\delta$ in the $\epsilon,\delta$ definition of continuity. This will help greatly to allay confusions. – Lee Mosher Mar 9 at 19:54
• @ Lee Mosher: I changed it to $\delta'$. Now, can you explain why such a $\delta'$ exist? – Sepide Mar 9 at 20:01

## 1 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 $$A$$.

• What I meant as this point was at this stage, now can you explain why we can go backward? – Sepide Mar 9 at 19:56
• Can you elaborate on what you mean by "going backwards"? – rolandcyp Mar 9 at 19:57
• Are you asking how we can deduce the continuity of $f$ at $x_0$? Or how you can deduce the continuity on all of $A$? – rolandcyp Mar 9 at 19:58
• $\delta$ is used for Lipschitzness of function and $\delta'$ is for continuity, can you explain question 1 first and say why such a $\delta'$ exist? – Sepide Mar 9 at 20:03
• See my edit, let me know if anything is still confusing. – rolandcyp Mar 9 at 20:10