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: You can just take the coefficients of the polynomials in $E$ to have $E$-valued polynomials. Trying to mimic the Stone-Weierstraß proof for $E$-valued functions may however be tricky.
But from uniform continuity of continuous functions on $[0,1]$ you immediately get that you can uniformly approximate all functions in $C([0,1],E)$ by piecewise affine functions. A piecewise affine function has its values in a finite-dimensional subspace of $E$, so the straightforward generalisation of Stone-Weierstraß to $\mathbb{R}^n$-valued functions gives the result.
A: Seems like the answer is yes. Take $f\in C([0,1],E)$. First extend $f$ to a function in $C(\Bbb R, E)$, say by making $f$ constant on $[1,\infty)$ and constant on $(-\infty,0]$. Now say $\phi_n$ is a smooth (real-valued) approximate identity; then the convolution $f*\phi_n$ should be differentiable and it should be that $f*\phi_n\to f$ uniformly on $[0,1]$. The proofs of those last two assertions should be just the same as for a real-valued function $f$, seems to me...
Yes. As above, assume that $f:\Bbb R\to E$ is bounded and uniformly continuous. Say $\phi_n\ge0$, $\int \phi_n=1$, $\phi_n$ is supported in $(-1/n,1/n)$ and $\phi_n$ is continuously differentiable. Let $$f_n(x)=f*\phi_n(x)=\int f(t)\phi_n(x-t)\,dt=\int f(x-t)\phi_n(t)\,dt.$$(There are no problems with any of the integrals here or below since we're integrating continuous $E$-valued functions with compact support; in fact we could take all the integrals to be Riemann integrals...)
First, $$||f_n(x)-f(x)||=\left|\left|\int(f(x-t)-f(x))\phi_n(t)\right|\right|\le\int_{-1/n}^{1/n}||f(x-t)-f(x)||\phi_n(t)
\le\sup_{|s-t|<2/n}||f(s)-f(t)||,$$so that $f_n\to f$ uniformly on $\Bbb R$.
Now to show that $f_n$ is continuously differentiable. Going to write this part in a little less detail. In fact it turns out that $$f_n'=f*(\phi_n'),$$which shows that $f_n'$ is continuous by an argument like the above, since $f$ is uniformly continuous and $\phi_n'$ is integrable. Note that $$\frac{f_n(x+h)-f_n(x)}{h}=\int f(t)\frac{\phi_n(x+h-t)-\phi_n(x-t)}{h}\,dt:=\int f(t)\psi_h(x-t)\,dt$$where $$\psi_h(t)=\frac{\phi_n(t+h)-\phi_n(t)}{h}.$$The hypotheses on $\phi_n$ imply that $$\lim_{h\to0}\int|\psi_h(t)-\phi_n'(t)|\,dt=0,$$and so again, since $f$ is bounded, inequalities like the above show that $$\lim_{h\to0}\frac{f_n(x+h)-f_n(x)}{h}=f*(\phi_n')(x).$$
A: 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.
