# Prove that Locally Lipschitz on a Compact Set implies Lipschitz

• Only a Mirage
In summary, we are trying to prove that a function that is locally Lipschitz on a compact subset W of a metric space M is Lipschitz on W. We can do this by assuming that the function is not locally Lipschitz, then extracting convergent subsequences using compactness. This leads to a contradiction, proving that the function must be Lipschitz on W.
Only a Mirage

## Homework Statement

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.

## Homework 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 $L_i$.

## 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

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##?

1 person
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.

1 person
LCKurtz said:
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##?

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!

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micromass said:
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.

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:

## 1. What is the definition of "Lipschitz continuity"?

Lipschitz continuity is a mathematical concept that describes how a function's rate of change is limited by a constant factor. A function f is said to be Lipschitz continuous if there exists a positive constant K such that the absolute value of the difference between the function's outputs at any two points is less than or equal to the constant multiplied by the distance between the points. In other words, the function's slope or rate of change is always bounded by this constant value.

## 2. What does it mean for a function to be "Locally Lipschitz"?

A function is said to be locally Lipschitz if it is Lipschitz continuous on every compact subset of its domain. This means that on any bounded, closed interval, the function's rate of change is limited by a constant value. However, the constant value may vary on different subsets of the domain.

## 3. What is a "compact set"?

A compact set is a subset of a metric space that is closed and bounded. In simpler terms, it is a set where every sequence of points in the set has a limit that is also within the set. In the context of proving Lipschitz continuity, we are specifically interested in compact sets that are subsets of the function's domain.

## 4. How does being locally Lipschitz on a compact set imply Lipschitz continuity?

When a function is locally Lipschitz on a compact set, it means that there exists a constant value K that bounds the function's rate of change on every compact subset of its domain. This constant value K can then be used as the Lipschitz constant for the entire function, making it Lipschitz continuous. This is because any two points in the function's domain can be connected by a sequence of compact subsets, and the Lipschitz constant on each subset will ensure that the function's rate of change is limited on the entire domain.

## 5. How is the proof of this statement typically presented?

The proof usually involves breaking down the function's domain into a sequence of nested compact subsets and using the Lipschitz constant on each subset to show that the function's rate of change is bounded on the entire domain. It may also involve using the Lipschitz constant and the definition of Lipschitz continuity to construct a sequence of points that converge to the desired Lipschitz constant.

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