Re: Uniform Lipschitz continuity


David C. Ullrich wrote:
On 5 Jul 2006 09:54:00 -0700, "dotVleq0" <dotVleq0@xxxxxxxxx> wrote:


David C. Ullrich wrote:
On 4 Jul 2006 12:24:52 -0700, "dotVleq0" <hfgrip@xxxxxxxxx> wrote:


David C. Ullrich wrote:
On 4 Jul 2006 02:46:20 -0700, "dotVleq0" <dotVleq0@xxxxxxxxx> wrote:

I am trying to find out whether the following holds:

If f(t,x) is locally Lipschitz continuous in x, uniformly in t on
RxR^n, then for any compact subset D of R^n, it is Lipschitz continuous
in x, uniformly in t on RxD.

I can imagine various things that various of those modifiers
might mean. Try restating the problem without using the
words "locally" or "Lipschitz", spelling out precisely
what the hypothesis and the conclusion are (via
inequalities and quantifiers).

Ok, I'll try writing it out more precisely. I would like to prove that
if the following statement is true for a function f:RxR^n->R^m:

for each x0 in R^n there exists a constant L and a neighborhood N
around x0 such that for all (t,x,y) in RxNxN |f(t,x)-f(t,y)| <= L|x-y|.

then the following holds:

For each D which is a compact subset of R^n there exists a constant L
such that for all (t,x,y) in RxDxD |f(t,x)-f(t,y)| <= L|x-y|.

(|*| denotes any p-norm). Does that resolve the ambiguities?

Yes, it's now clear what the question is (and no, this is not
what I thought the question was.) This statement is true
(proof below).

Thank you for a very good reply. If you don't mind, I just have a
couple of follow-up questions (below). Please feel free to ignore them
if they don't make sense.


I think it comes down to whether D can be covered by a finite number of
the neighborhoods N.

Two comments on that:

(i) Yes, it's true that if D is compact and D is covered by a bunch
of open sets N then D is covered by finitely many of the N. This
is very well-known, in fact in general topology it's the _definition_
of "D is compact".

What I started wondering was whether it was possible for some functions
(with the "local Lipschitz" property that I defined) that the sets N
would have to be made infinitely small somwhere (approaching a point).
It seems intuitively logical that this could not be the case on a
compact subset of R^n, but I haven't been able to conclusively prove
that for myself. I realise it may seem like a silly question, but for
someone not well-versed in toplogy, can you give a hint about what form
such a proof would take (if it is in fact needed)?

(ii) Actually the proof depends on more than that. The proof I have
in mind uses the fact that any compact subset of R^n is contained
in some convex compact subset of R^n (for example a closed ball).

Example showing that it doesn't follow just from the fact that
any open cover of a compact set has a finite subcover:

Say X = (0,1) union (2,3), the union of two intervals. Consider
these two statements:

(a) for each x0 in X there exists a constant L and a neighborhood
N around x0 such that for all (t,x,y) in RxNxN |f(t,x)-f(t,y)|
<= L|x-y|.

(b) For each D which is a compact subset of X there exists a
constant L such that for all (t,x,y) in RxDxD |f(t,x)-f(t,y)|
<= L|x-y|.

Here (a) does _not_ imply (b), even though what you say about
N's and D's is still true. For example, define f(t,x) = t
for x in (0,1), f(t,x) = 2t for x in (2,3). Then (a) is
true and (b) is false.

Just out of curiosity, if we consider only the case when X = R^n (as
was the premise in the original statement) are there still cases where
(a) does not imply (b)?

Huh? Did you read the whole post?

I did, but I clearly misread this part. I don't know quite what I was
thinking--sorry about that.



This is stated a mere fact in a couple of proofs I
have seen for the case when f does not depend on t. If anyone can point
me to a proof of why this is, I would greatly appreciate it. I can see
no reason why it would be different in this case, but I would like to
confirm it.

How you prove that fact about compact sets depends on what definition
of "compact" you're using.

Compact (closed and bounded) subsets of R^n are the ones that I am
interested in in this case. Are there several possible definitions
here?

Presumably you've proved that any sequence in a closed and bounded
set has a convergent subsequence. That means your closed and bounded
set is a "sequentially compact" metric space. The fact that a
sequentially compact metric space is actually compact (ie, every
open cover has a finite subcover) is something you can find in
a lot of places.

Let's see, for a closed and bounded subset of R^n some of the details
may be simpler. Let's say D is a closed and bounded subset of R^n,
we have a collection of open set N_j, and D is contained in the union
of the N_j. We need to show that D is contained in the union of
finitely many of the D_j.

The first thing to show is that there exists a "Lebesgue number":

(*) There exists r > 0 with the property that for every x in D there
exists a set N_j in our collection such that B(x,r) is a subset
of N_j.

Here B(x,r) is the open ball with center x and radius r. You need
to read (*) very carefully; people sometimes get the idea that it
says more than it does.

Proof of (*): Suppose that (*) is false. There does not exist
r > 0 such that etc. Then for each n, r =1/n does not work,
so there exists x_n in D such that B(x_n, 1/n) is not contained
in any of the sets N_j.

Now the sequence x_n has a convergent subsequence. Just to make
this easier to type, let's pretend that the sequence x_n itself
is convergent: x_n -> x. Then x is in D since D is closed.
So there exists a set N = N_j with x in N. Since N is open
there exists r > 0 such that B(x,r) is a subset of N.

Now choose n so large that two things happen: |x_n - x| < r/2
and 1/n < r/2. It follows from the triangle inequality that
B(x_n, 1/n) is a subset of B(x,r). Hence B(x_n, 1/n) is
a subset of N, contradicting the assumption that it is
not a subset of any N_j. QED(*)

Now I get it. The existence of a convergent subsequence was the missing
link that didn't occur to me.

Thank you for the excellent reply!


Now let S be the set of all balls of the form B = B(x,r/3)
such that B is a subset of some N_j. The balls in S cover D.
Let F be a maximal disjoint subcollection of S: F is a subset
of S, the balls in F are pairwise disjoint, and any ball
in S intersects some ball in F. (You can construct F just
by starting with any element of S. If there is another
element of S disjoint from the first element of F, throw
it into F. Continue this way - at each stage, if there
is a ball in S disjoint from all the currect elements of
F, pick one and add it to F. When you run out of balls
in S to add to F it must be because every remaining ball
in S intersects some element of F.)

Then F must be finite, because the balls in F are disjoint,
they have a fixed size, and their union is bounded. Given
B = B(x,r) in F let B' = B(x,r). Then in fact the balls
B' for B in F cover D. (Because if x is any point of D then
B(x,r/3) is an element of S. So there exists B(y,r/3)
in F which intersects B(x,r/3), and then the triangle
inequality shows that B(x,r/3) is a subset of B(y,r),
which is one of the B'.) But each B' is a subset of some
N_j, so there are finitely many N_j that cover D.


Anyway, here's a proof that your assertion is true:

Since any compact D is contained in a closed ball, we can
assume that D is itself a closed ball. Now D is in fact
covered by finitely many of those N's; let L be the
largest of the L's associated with each of the finitely
many N's that cover D.

Say x, y are in D. Draw a straight line from x to y. Choose
points

x = x_0, .... x_k = y

such that the x_j lie in order along that line, and such
that each x_j is so close to its successor x_{j+1} that
x_j and x_{j+1} lie in one of the N's. Then

|f(t,x_j) - f(t,x_{j+1}| <= L|x_j - x_{j+1}|

for every j, and so the triangle inequality shows that

|f(t,x) - f(t,y)| <= L sum|x_j - x_{j+1}| = L|x-y|.

************************

David C. Ullrich

Thanks again!


************************

David C. Ullrich

.

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