# Prove that Locally Lipschitz on a Compact Set implies Lipschitz

1. Mar 16, 2014

### Only a Mirage

1. The problem statement, all variables and given/known data
Let $M$ and $N$ be two metric spaces. Let $f:M \to N$. Prove that a function that is locally Lipschitz on a compact subset $W$ of a metric space $M$ is Lipschitz on W.

A similar question was asked here

but it didn't really address my question.

2. Relevant equations
Definition: a function $f:M \to N$ is said to be Lipschitz on a set $S$ if there exists a positive constant $L \in \Re^+$ such that $\forall x,y \in S$, $d_N(f(x),f(y)) \leq L*d_M(x,y)$

Definition: a function $f:M \to N$ is said to be locally Lipschitz on a set $S$ if for every point $x_i \in S$, there exists an open ball $B_{r_i}(x_i)$ of radius $r_i$ centered at $x_i$ such that $f$ is Lipschitz on $B_{r_i}(x_i)$ with Lipschitz constant $L_i$.

3. The attempt at a solution
Because we are given that $f$ is locally Lipschitz, we know that $\forall x_i \in W: \exists r_i \in \Re^+: f$ is Lipschitz on $B_{r_i}(x_i)$. The set $\{B_{r_i}(x_i) | x_i \in W \}$ is an open cover of $W$. Since $W$ is compact, we can extract a finite subcover so that (after a possible re-ordering of indices of balls) $$W \subseteq \bigcup_{i=1}^{N}B_{r_i}(x_i)$$.

Now, for any two $x,y \in W$ there are two possibilities:
i) $\exists i \in \{1,...,N\}: x,y \in B_{r_i}(x_i)$. In this case, we have $d_N(f(x),f(y)) \leq L_i*d_M(x,y)$. If this were true for all pairs of points, I could simply choose a Lipschitz constant for $W$ to be $\max_i(L_i)$.

ii) $\forall i \in \{1,...,N\}: x, y$ are not in the same ball $B_{r_i}(x_i)$. In this case, I'm really stuck regarding what to do. I have tried using the triangle inequality with little success.

This actually isn't a homework assignment. The ODE textbook I'm using stated this fact, but relegated its proof to the exercises. Any help would be appreciated

2. Mar 16, 2014

### LCKurtz

Say, for example that $x$ and $y$ happen to be in distinct overlapping balls $B_i$ and $B_j$. Pick $z\in B_i\cap B_j$. Then can you find an $L$ that works by considering the sequence of points $x, x_i, z, x_j, y$?

3. Mar 16, 2014

### micromass

I'm not saying an open cover argument won't work here, but it'll be annoying. I prefer to work with the sequential definition of compactness here.

So, I would do the following: Assume that $f$ is not locally Lipschitz, then there exists sequences $(x_n)_n$ and $(y_n)_n$ such that

$$d(f(x_n),f(y_n))> n d(x_n,y_n)$$

for each $n$. Now apply compactness to extract convergent subsequences.

4. Mar 16, 2014

### Only a Mirage

I was thinking along these lines earlier. Then $$d_N(f(x),f(y)) \leq d_N(f(x),f(z)) + d_N(f(z),f(y)) \leq L_1*d_M(x,z) + L_2*d_M(z,y)\\ \leq \max(L_1,L_2)*(d_M(x,z)+d_M(z,y))$$

But I'm stuck here. How would I use $x_i$ and $x_j$? Also, how can I show that there is always a "chain" of overlapping balls to get from $x$ to $y$? Do I need to assume $W$ is connected?

Another thing I tried is to try to bound from below $d_M(x,y)$ given that $x,y$ are not in the same ball (My book gives a hint along these lines, but I think the hint is incorrect). If I can show that in this case $\exists k \in \Re^+: k \leq d_M(x,y)$,
then since f is a continuous map, by compactness its image is bounded. Therefore, $$\exists C \in \Re^+: d_N(f(x),f(y)) \leq C \leq \frac{C}{k}d_M(x,y)$$

And we would be done since we could take $$L = \max \{L_1,...,L_N,\frac{C}{k}\}$$However, I haven't been able to find such a $k$. It would be satisfying for me to be able to complete this problem using this covering argument!

Last edited: Mar 16, 2014
5. Mar 16, 2014

### Only a Mirage

Interesting. I'd never heard of sequential compactness before, but I just found some notes online and read about it. Did you mean to say "assume $f$ is not Lipschitz"? We are given in the problem that $f$ is locally Lipschitz, so I don't know why you would assume otherwise.

If I'm right in thinking that you meant to say "assume $f$ is not Lipschitz", then okay: so let's say I extract convergent subsequences $x_{n_k}\to x$ and $y_{n_k} \to y$. Then we have $d(f(x_{n_k}),f(y_{n_k})) > n_k* d(x_{n_k},y_{n_k})$

Since by continuity of $f$ and compactness of $W$, $f(W)$ is a bounded set. Hence, $d(f(x_{n_k}),f(y_{n_k}))$ is bounded. Therefore, we must have $d(x_{n_k},y_{n_k}) \to 0$, so in fact $x = y$.

Now consider an arbitrary $\epsilon \in \Re_{>0}$. We now know that $\exists K \in \Re_{>0}: \forall k \geq K: x_{n_k},y_{n_k} \in B_{\epsilon}(x)$. But for any positive constant $m$, $\exists k: n_k > m$ and $d(f(x_{n_k}),f(y_{n_k})) > n_k* d(x_{n_k},y_{n_k})$. Therefore, in every $\epsilon$ ball centered at $x$, $f$ is not Lipschitz, which by definition implies that f is not locally Lipschitz. Since this contradicts the hypothesis, it must be the case that $f$ is Lipschitz on $W$.

Does this argument sound correct? I'd still like to be able to prove the theorem using my covering argument (only because I'm stubborn), but thank you for your help and for teaching me something useful (sequential compactness) .

Last edited: Mar 16, 2014