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.

$\begingroup$ 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? $\endgroup$ – Amir Sagiv Oct 21 at 0:46

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4$\begingroup$ @AmirSagiv: A Banach space with this property is usually said to have the RadonNikodym property, and for instance every reflexive Banach space has this property. I added a few details in an additional answer below. $\endgroup$ – Jochen Glueck Oct 21 at 6:21

$\begingroup$ 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. $\endgroup$ – 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 RadonNikodym 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 RadonNikodym property [1, Corollary 1.2.7].
If $X$ is separable and has a predual Banach space, then $X$ has the RadonNikodym property [1, Theorem 1.2.6].
If $V$ is a closed vector subspace of $X$ and $X$ has the RadonNikodym property, then so has $V$ (this is obvious, of course).
The space $c_0$ does not have the RadonNikodym 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 RadonNikodym property.
In particular, the space $C([0,1])$ does not have the RadonNikodym 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 RadonNikodym property [1, Proposition 1.2.10].
Reference:
[1]: Arendt, Batty, Hieber, Neubrander: "Vectorvalued 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 RadonNikodym property, then there always exists a Lipschitz continuous function $f: \mathbb{R} \to X$ that is nowhere differentiable.

2$\begingroup$ 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. $\endgroup$ – Anthony Quas Oct 21 at 6:39

1$\begingroup$ 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. $\endgroup$ – Bill Johnson Oct 22 at 17:53

1$\begingroup$ @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 RadonNikodym property". The space $\ell^1$ has the RadonNikodym property, so if the second bullet point were correct as written, it would imply that $c_0$ had the RadonNikodym property, which it doesn't. $\endgroup$ – Robert Furber Oct 22 at 20:09

$\begingroup$ @RobertFurber: Oops, you're right, of course! Thanks a lot, I've corrected the statement. $\endgroup$ – Jochen Glueck Oct 22 at 20:15

$\begingroup$ @BillJohnson: Thanks a lot, it is good to know this. I will include your comment in the answer. $\endgroup$ – Jochen Glueck Oct 22 at 20:17