Friday, May 6, 2016

real analysis - Continuous or Differentiable but Nowhere Lipschitz Continuous Function




  1. What is a real valued function that is continuous on a close interval but not Lipschitz continuous on any subinterval?



  2. What is a real valued function that is differentiable on a close interval but not Lipschitz continuous on any subinterval?



Answer



Continuous and nowhere Lipschitz



An example is given by the Weierstrass function, which is continuous and nowhere differentiable. This can be justified in two ways:




  • A Lipschitz function is differentiable almost everywhere, by Rademacher's theorem.


  • Direct inspection of the proof that the function is nowhere differentiable; the estimates used in the proof also imply it's nowhere Lipschitz ($|f(x+h)-f(x)|$ is estimated from below by $|h|^\alpha$ with $\alpha<1$, for certain $x,h$).





Differentiable and nowhere Lipschitz



There are no such examples: a differentiable function on an interval must be Lipschitz on some subinterval. The following is an adaptation of a part of PhoemueX's answer.




  1. The function $f'$ is a pointwise limit of continuous functions: namely, $$f'(x) = \lim_{n\to\infty} n(f(x+1/n)-f(x))$$
    where for each $n$, the expression under the limit is continuous in $x$.


  2. Item 1 implies that $f'$ is continuous at some point $x_0$. This is a consequence of a theorem about functions of Baire class 1: see this answer for references.



  3. Continuity at $x_0$ implies $f'$ is bounded on some interval $(x_0-\delta,x_0+\delta)$. The Mean Value Theorem then implies that $f$ is Lipschitz on this interval.



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...