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  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?

Hans
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    so the derivative has to be unbounded on any neighborhood. maybe look at $\sqrt{1-x^2}$ odd symmetrized and periodized, call this $h(x)$. it's derivative is unbounded at $x=0$. then look at $g(x) = \sum_{n=1}^\infty c_n h(nx)$ for some (summable) sequence $(c_n)$, – reuns May 26 '16 at 09:18

1 Answers1

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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.

Community
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  • Thank you, sandwich. I will study your answer to part 2). Meanwhile, could you please think about part 1) for $f$ being continuous? I have now edited the question and separated it explicitly into two distinct parts. I should have done that at the beginning to prevent the confusion. – Hans May 26 '16 at 19:32
  • Okay, I expanded the answer. –  May 26 '16 at 19:48