# Lipschitz functions

## Homework Statement

1. Let 0 < a < b <= 1. Prove that the set of all Lipschitz functions of order
b is contained in the set of all Lipschitz functions of order a.

2. Is the set of all Lipschitz functions of order b a closed subspace of those
of order a?

## Homework Equations

I know that a function f: [a,b] -> R is Lipschitz of order a if there exists a constant K
such that |f(x) - f(y)| <= K |x-y|^a and for all x,y in [a,b].

## The Attempt at a Solution

Assume f is a Lipschitz function of order b then there exists some constant K such that
|f(x)-f(y)|<= K |x-y|^b. Then I need to prove that we can find some constant say C
such that |f(x) - f(y)| <= C |x-y|^a , where 0 < a < b=1.

Fix a and b. Have you considered the equality K|x - y|b = K|x - y|a|x - y|b - a ?

Fix a and b. Have you considered the equality K|x - y|b = K|x - y|a|x - y|b - a ?

So |f(x)-f(y)|<= K |x-y|^b implies |f(x)-f(y)|<= K |x-y|^a |x-y|^(b-a).

Therefore: |f(x) - f(y)| /|x-y|^a <= |x-y|^(b-a)

But x and y are both in [a,b] so |x-y| <= |x|+|y| = b + b = 2b.

Therefore |f(x)-f(y)|/|x-y|^a <= (2b)^(b-a).

So our constant C is then (2b)^(b-a). Is this OK?

How to show the closedness part? I know I have to take a sequence and show its closed
under the limit but really I have no clue how to proceed.

So |f(x)-f(y)|<= K |x-y|^b implies |f(x)-f(y)|<= K |x-y|^a |x-y|^(b-a).

Therefore: |f(x) - f(y)| /|x-y|^a <= |x-y|^(b-a)
What happened to K?

But x and y are both in [a,b] so |x-y| <= |x|+|y| = b + b = 2b.
The first equality does not make sense, x and y are variables. Add some more rigor to your statements.

OK, thanks again.
My try:

Since f is Lipschitz of order b then there exists a constant K >0 such that
for all x, y in [a,b] we have: |f(x)-f(y)|<= K |x-y|^b.

Observe K|x-y|^b = K|x-y|^a |x-y|^(b-a).

Therefore |f(x)-f(y)| <= K |x-y|^a |x-y|^(b-a) and thus:

|f(x)-f(y)|/|x-y|^a <= K |x-y|^(b-a).

Since x, y are points in [a,b] then |x-y|^(b-a) <= (2b)^(b-a).

Therefore |f(x)-f(y)|/|x-y|^a <= K (2b)^(b-a) and hence:

|f(x)-f(y)|<= K (2b)^(b-a) |x-y|^a so f is Lipschitz of order a with constant
C = K (2b)^(b-a).

OK?

HallsofIvy
Homework Helper
OK, thanks again.
My try:

Since f is Lipschitz of order b then there exists a constant K >0 such that
for all x, y in [a,b] we have: |f(x)-f(y)|<= K |x-y|^b.

Observe K|x-y|^b = K|x-y|^a |x-y|^(b-a).

Therefore |f(x)-f(y)| <= K |x-y|^a |x-y|^(b-a) and thus:

|f(x)-f(y)|/|x-y|^a <= K |x-y|^(b-a).

Since x, y are points in [a,b]
You appear to be using a and b with two different meanings here. I think more important is that since a< b< 1, 0< b-a< 1.

then |x-y|^(b-a) <= (2b)^(b-a).

Therefore |f(x)-f(y)|/|x-y|^a <= K (2b)^(b-a) and hence:

|f(x)-f(y)|<= K (2b)^(b-a) |x-y|^a so f is Lipschitz of order a with constant
C = K (2b)^(b-a).

OK?

Halls: Sorry, I do not follow your hint/suggestion, what do you mean?

Halls: Sorry, I do not follow your hint/suggestion, what do you mean?

Two things that Hall mentioned:
1) You are denoting the endpoints of the closed interval with the same constants you are using to denote the exponents of the Lipschitz inequality. The two are not related; if necessary use different letters, like [c, d].
2) The fact that 0 < b-a < 1 is important.

Two things that Hall mentioned:
1) You are denoting the endpoints of the closed interval with the same constants you are using to denote the exponents of the Lipschitz inequality. The two are not related; if necessary use different letters, like [c, d].
2) The fact that 0 < b-a < 1 is important.

Gotcha guys.

|f(x)-f(y)| / |x-y|^a<= K |x-y|^(b-a).

Since 0 < b-a< 1 then |x-y|^(b-a) < |x-y|.

Actually I meant: then |f(x)-f(y)|/|x-y|^a <= K |x-y|^a |x-y|. But x,y are both points in [c,d] so |x-y| <= d.

Thus we get |f(x)-f(y)| <= K |x-y|^a * d so the constant is C = k*d, correct?

Last edited:
Well since a and b are endpoints, and x,y are any points in [a,b], the inequality |x-y| =< |b -a| holds, so I don't think you need any extra variables beyond what's given in the problem statement.