# Non-differentiable Lipschitz functions

As far as I understand, there are Lipschitz functions $$f:\mathbb{R}\to\ell^\infty$$ that are nowhere differentiable in the Frechet sense. Where can I find such an example?

Define $$f_n(t)=\sin(nt)/n$$ and $$\phi(t)=(f_1(t),\ldots)$$. This has Lipschitz constant 1, but has no Fréchet derivative as far as I can tell.

• Interesting. So what would be a sufficient criterion for a space to have a "Rademacher Theorem", i.e., that Lipschitz functions are almost everywhere differentiable? – Amir Sagiv Oct 21 at 0:46
• Could Finite dimensionality be necessary? – Anthony Quas Oct 21 at 0:46
• @AmirSagiv: A Banach space with this property is usually said to have the Radon-Nikodym property, and for instance every reflexive Banach space has this property. I added a few details in an additional answer below. – Jochen Glueck Oct 21 at 6:21
• An account of differentiability of Banach space valued Lipschitz functions of a real variable is given in Section 6.1 on pages 111−114 in S. Yamamuro's Differential Calculus in Topological Linear Spaces, Springer LNM 374, 1974, There are some sufficient conditions for a Lipschitz function to be a.e. differentiable. In particular, being finite dimension is not necessary. – TaQ Oct 21 at 7:52

In addition to Anthony Quas' answer, it might be worthwhile to mention the following general observations.

A Banach space $$X$$ is said to have the Radon-Nikodym property if every Lipschitz mapping $$f: \mathbb{R} \to X$$ is differentiable almost everywhere.

Here are a some interesting observations concerning this property (which can all be found in [1, Section 1.2]):

• Every reflexive Banach space has the Radon-Nikodym property [1, Corollary 1.2.7].

• If $$X$$ is separable and has a pre-dual Banach space, then $$X$$ has the Radon-Nikodym property [1, Theorem 1.2.6].

• If $$V$$ is a closed vector subspace of $$X$$ and $$X$$ has the Radon-Nikodym property, then so has $$V$$ (this is obvious, of course).

• The space $$c_0$$ does not have the Radon-Nikodym property [1, Proposition 1.2.9]. (The counterexample given there is actually the same as in Anthony Quas' answer). It follows in particular that a Banach space that contains an isomorophic copy of $$c_0$$ as a closed subspace, does not have the Radon-Nikodym property.

• In particular, the space $$C([0,1])$$ does not have the Radon-Nikodym property. A more explicit counterexample on this space can also be found in [1, Example 1.2.8(a)].

• The space $$L^1(0,1)$$ does not have the Radon-Nikodym property [1, Proposition 1.2.10].

Reference:

[1]: Arendt, Batty, Hieber, Neubrander: "Vector-valued Laplace Transforms and Cauchy Problems" (2011).

Remark The question asks for a Lipschitz function $$f: \mathbb{R} \to X$$ which is nowhere differentiable. As pointed out by Bill Johnson (in the comments below this answer), if a Banach space $$X$$ does not have the Radon-Nikodym property, then there always exists a Lipschitz continuous function $$f: \mathbb{R} \to X$$ that is nowhere differentiable.

• This is very interesting. I sort of suspected I could port the example to $\ell^2$, but I guess not... I should probably try and see what goes wrong. – Anthony Quas Oct 21 at 6:39
• A Banach space $X$ fails the RNP iff there is a nowhere differentiable Lipschitz function $f$ from $[0,1]$ into $X$ iff there is $\epsilon > 0$ and a Lipschitz function $f$ from $[0,1]$ into $X$ s.t. $f$ has no point of $\epsilon$-differentiablity. See e.g. Proposition 5.21 in the book of Benyamini and Lindenstrauss. Of course the unit interval can be replaced by the real line in the equivalences. – Bill Johnson Oct 22 at 17:53
• @JochenGlueck The second bullet point is not correct (and does not represent Theorem 1.2.6 correctly). The correct statement is "If $X$ is the dual space of a Banach space and $X$ is separable, then $X$ has the Radon-Nikodym property". The space $\ell^1$ has the Radon-Nikodym property, so if the second bullet point were correct as written, it would imply that $c_0$ had the Radon-Nikodym property, which it doesn't. – Robert Furber Oct 22 at 20:09
• @RobertFurber: Oops, you're right, of course! Thanks a lot, I've corrected the statement. – Jochen Glueck Oct 22 at 20:15
• @BillJohnson: Thanks a lot, it is good to know this. I will include your comment in the answer. – Jochen Glueck Oct 22 at 20:17