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Lipschitz Continuous Functions

Date: 04/27/2008 at 08:21:14
From: Bel
Subject: Lipschitz continuous functions

(a)Show that a Lipschitz continuous function is uniformly continuous 
on a subset and that the converse is not necessarily true.  
(b) Give an example to show that Lipschitz continuous functions are 
not necessarily differentiable.

I am aware of the theorem of Lipschitz but I am just not sure how to 
begin my proof for a.  So, Lipschitz tells us that between no two 
points of a function the slope of the gradient will never be any 
bigger than that constant L.  I really have no idea where to begin my 
proof or how to carry it out.

Lipschitz denotes the "smoothness" of the curve I suppose.  Maybe if 
I start using given epsilon > 0 or something?



Date: 04/27/2008 at 17:44:40
From: Doctor Jordan
Subject: Re: Lipschitz continuous functions

Hi Bel,

For (a), the first problem is to show that if f:D->R is Lipschitz
continuous then it is uniformly continuous.  When we are stuck it is
helpful to write out the definitions.

f being Lipschitz continuous on D means there is a constant K>0 such
that for all x,y in D, 

  |f(x)-f(y)| <= K|x-y| where <= means less than or equal.

For f to be uniformly continuous on D means that for all epsilon > 0,
there is a delta > 0 such that

  |x-y| < delta implies |f(x)-f(y)| < epsilon.

We want to show that f is uniformly continuous on D.  Let epsilon > 0.
Now let delta = epsilon/K.

If x,y are in D and |x-y| < delta, then

  |f(x)-f(y)| <= K|x-y| since f is Lipschitz
               < K*delta since |x-y| < delta
               < K*epsilon/K since delta = epsilon/K
               < epsilon,

hence |f(x)-f(y)| < epsilon, thus f is uniformly continuous on D.  
This shows that f being Lipschitz continuous implies that f is 
uniformly continuous.

For the second part of (a) we want to find a function that is
uniformly continuous but not Lipschitz continuous.

The function f(x) = sqrt(x) is continuous on the closed interval
[0,1].  Since [0,1] is compact, f is uniformly continuous on [0,1]. 
It is a fact that a continuous function on a compact set is uniformly
continuous.  If you have not seen this before, check your book or look
online, but if you still don't understand it write me back about that.

We will show that f(x) is not Lipschitz continuous on [0,1].  If f
were Lipschtz with constant K, then for all x in [0,1] we would have

  |f(x)-f(0)| <= K|x-0|

i.e.

  |sqrt(x)-sqrt(0)| <= K|x|

i.e.

  |sqrt(x)| <= K|x|

i.e. for all x in (0,1] (we exclude 0 so we can divide)

  1/K <= |sqrt(x)| by dividing both sides by |sqrt(x)|

But 1/K is a fixed constant.  For

        1
  x = -----
      4*K^2,

then sqrt(x) = 1/(2*K), which implies

  1/K <= 1/(2*K),

a contradiction. Therefore f is not Lipschitz continuous on [0,1].

Please check this over carefully but write me back if it doesn't make
sense.

For (b), define f:R->R by f(x) = |x|.  It is a fact that f is not
differentiable on all of R, since it is not differentiable at the
point x = 0.

The triangle inequality tells us that |a+b| <= |a| + |b| for all a,b
in R.  Therefore 

  |a| >= |a+b| - |b|,

so

  |x-y| >= |x-y+y| - |y|,

so

  |x-y| >= |x| - |y| for all x,y in R.

But then also |y-x| >= |y| - |x| for all x,y in R. 

Therefore

  |x-y| >= | |x| - |y| |

for all x,y in R.

Now, let K=1.  For all x,y in R,

  |f(x)-f(y)| = | |x| - |y| | <= |x-y| = K*|x-y|,

hence f is Lipschitz continuous on R with Lipschitz constant K=1. 
This gives an example of a function that is Lipschitz continuous but
not differentiable.

- Doctor Jordan, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Functions

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