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I am looking for examples of functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that:

  • are differentiable, and
  • have discontinuous but bounded derivatives.

I believe that a function satisfying these conditions must also be (globally) Lipschitz continuous.

The only example I know of is $f(x) = \left\{ \begin{array}{c l} x^2\cdot \sin\left(\frac{1}{x}\right) & ,\quad x\neq0\\ 0 & ,\quad x=0 \end{array} \right.$, so examples besides this one would be most helpful.

My question is inspired by having recently proven that if a differentiable function $f$ has a bounded derivative, then $f$ is Lipschitz. A differentiable function with a continuous derivative is also Lipschitz, but requiring $f'$ to be continuous is not necessary for Lipschitz continuity. This exercise made me curious about examples of differentiable functions that meet the bounded derivative requirement but not the (stricter) continuous derivative requirement.

Thanks in advance for any suggestions!

Math1000
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qwerty
  • 23

1 Answers1

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An example is $f(x)=x^2\sin(1/x)$. Your claim is true, since for example Lagrange theorem doesn't require the derivative to be continous, and then the bound on the derivative is a suitable Lipschitz constant.

Lorenzo Cecchi
  • 442
  • Thanks, Lorenzo! Do you know of any other examples? That's the only one I've been able to identify. (Besides the cosine version of the same function...) – qwerty Nov 01 '19 at 00:29
  • At the moment I have not anyone, but consider that you need to have $\liminf f'(x)\leq f'(0)\leq\limsup f'(x) $ (obviously 0 plays no particular role) – Lorenzo Cecchi Nov 01 '19 at 00:45