Is $C^1([0, 1], E)$ dense in $C([0, 1], E)$ for a general Banach-space $E$? I'm wondering if there is a result stating that the space of differentiable functions $C^1([0, 1], E)$ is dense in $C([0, 1], E)$ for a general Banach-space $E$. We define the derivative of a function $f : [0, 1] \rightarrow E$ simply as $$f'(t) := \lim_{h \rightarrow 0} \frac{f(t + h) - f(t)}{h}$$
I suspected this, since this is a Corollary from Stone-Weierstrass for $E = \mathbf{R}$. But of course one can't make generally sense of polynomials in Banach-spaces.

A very elegant proof can be based on tensor product representation
$$C([0,1],E)= C([0,1]) \tilde \otimes_\varepsilon E.$$
For every dense subspace $L$ of $C([0,1])$ the tensor product $L\otimes E$ is dense.
