Non-differentiable Lipschitz functions

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.

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.