I would like to solve the following exercise but there are a few minor things I am not clear about:
Let $A$ be the Banach algebra of $C^1([0,1])$ endowed with the norm $\|f\|=\|f\|_\infty + \|f'\|_\infty$. Let $p:[0,1]\to \mathbb C$ be the map $p(x) = x$. Show that $p$ generates $A$ as a Banach algebra.
Now, I believe generates as a Banach algebra means that the closure of the algebra generated by $p$ equals $A$.
Now question 1: When saying a function from a real interval into the complex numbers is differentiable, does it mean complex or real differentiable? I thought that it must mean complex differentiable. Is that so?
My next thought was that maybe the exercise is missing the unit. If the set was $\{1,p, \overline{p}\}$ then it would follow from the Stone Weierstrass theorem that this generates $C[0,1]$.
I guess if it is complex differentiable then it's analytic so can be expressed as a power series. Somehow that still needs a unit though. Or doesn't it?
The domain of the functions - $[0,1]$ - is real, so the differentiability is real differentiability.
Concerning the unit, the exercise presumes the generated algebra to be unital, so the Banach algebra generated by $p$ is the smallest closed subalgebra of $C^1([0,1])$ containing $1$ and $p$. (Otherwise, $p$ would not generate the full algebra.)
Note that $\overline{p} = p$ here, so the algebra generated by $p$ (or $1$ and $p$) is closed under conjugation. [You could also just look at only real-valued functions, it is an essentially real situation.]
The Stone-Weierstraß theorem deals with approximations in the $\lVert\cdot\rVert_\infty$ norm, which is different from the norm used here, so you cannot quite directly use it. However, a small indirection employs the Stone-Weierstraß theorem to get the result.