Local Lipschitz function with compact support

Question

Let $f:\mathbb{R}\to \mathbb{R}$ a locally Lipschitz continuous with compact support. Is it possible to say that $f$ is Lipschitz continuous over $\mathbb{R}$? over the set where $f$ is nonzero? It's a fact that if $f$ is locally Lipschitz over a compact $K$ then is Lipschitz over this compact. Is it posibble to conclude a similar fact here in this question? Please any help will be appreciated

Answer 1

$f$ is Lispchitz on the whole real line. First note that $f$ is bounded. Let $|f(x)| \leq C$ for all $x$. If $f$ is not Lipschtz then we can find points $x_n,y_n$ such that $|f(x_n)-f(y_n)| >n |x_n-y_n|$ for all $n$. This gives $2C >n |x_n-y_n|$ so $x_n-y_n \to 0$. Suppose $f(x)=0$ for $|x| >M$. Now $f$ is Lipschitz $[-M-1,M+1]$. If $|f(x)-f(y)| \leq K|x-y|$ on this interval you can easily check that $|f(x_n)-f(y_n)| \leq n |x_n-y_n|$ for all $n$ such that $|x_n-y_n| <1$ and $n>K$. This is a contradiction.

Answer 2

We can suppose that the support $S$ is a compact convex subset of $\mathbb{R}^n$.

First show that $f$ is lLc on $S$.

Since $f$ is lLc, for each $x$ there is some open $U$ and some $L$ such that $f$ is Lipschitz with rank $L$ on $U$. We can suppose that $U$ is convex without loss of generality.

Since $S$ is compact, it is covered by a finite number of open sets $U_1,...,U_m$ with corresponding ranks $L_1,..,L_m$.

Let $L = \max_i L_i$.

Choose $x,y \in S$ and let $p(t) = x+t(y-x)$ be the straight line joining $x,y$.

I need some notation: Call $[0,T] \subset [0,1]$ a finitely covered interval (FCP) iff there is a partition $t_0=0 < t_1 < \cdots < t_l = T$ such that each segment $[p(t_k),p(t_{k+1})]$ is contained in some $U_j$.

Let $M = \sup \{T \in [0,1] | [0,T] \text{ is a FCP} \}$. Since there is some $U_i$ with $p(0) = x \in U_i$ we see that $M >0$. Suppose $M < 1$. Since $p(M)$ is in some $U_q$, we see that there is some $\tau \in [0,M)$ such that $p(\tau) \in U_q$. Furthermore, there is some $M' \in (M,1]$ such that $p(M') \in U_q$. Since $[0,\tau]$ is a FCP (by definition of $T$) we see that $[0,M']$ is a FCP, which contradicts the definition of $T$. Hence $T=1$.

Finally, let $(t_0,...,t_l)$ be a partition of $[0,1]$ such that $[p(t_k),p(t_{k+1})]$ is contained in $U_{m_k}$, $k=0,...,l-1$.

Then $\|f(x)-f(y)\| \le \sum_{k=0}^{l-1} \|f(p(t_k)) - f(p(t_{k+1}))\| \le L \sum_{k=0}^{l-1} \|p(t_k) - p(t_{k+1})\| = L \|x-y\|$.

To finish, let $x,y$ be arbitrary and define $p$ as above.

If the segment $[x,y]$ does not intersect $S$ then we have $\|f(x)-f(y)\| = 0 \le L \|x-y\|$.

Otherwise, let $t_1 = \min \{ t \in [0,1] | p(t) \in S \} $ and $t_2 = \max \{ t \in [0,1] | p(t) \in S \} $. Since $S$ is convex, we see that $[p(t_1),p(t_2)] \subset S$.

Then \begin{eqnarray} \|f(x)-f(y)\| &\le& \|f(p(0))-f(p(t_1))\| + \|f(p(t_1))-f(p(t_2))\| + \|f(p(t_2))-f(p(1))\| \\ &=& \|f(p(t_1))-f(p(t_2))\\ &\le& L \|p(t_1)-p(t_2)\|\\ &\le& L \|p(0)-p(1)\|\\ &= & L \|x-y\| \end{eqnarray}