I want to show that the element $\exp( i \pi x)$ of $C([0,1])$ generates the entire algebra. My approach can be sketched as follows :
Let $f \in C([0,1])$. We consider the continuous even extension of $f$ given by $\tilde{f} : [-1,1] \to \mathbb{C}$.
Uniformly approximate $\tilde{f}$ with even polynomials. This is possible since the even polynomials satisfy the conditions of the Stone-Weierstrass theorem when we consider the algebra to be the algebra of even functions.
Even polynomials are twice continuously differentiable and $2$-periodic on the interval $[-1,1]$, so we can uniformly approximate each polynomial with its Fourier series. Since the interval is $[-1,1]$ the Fourier series of an even polynomial is given by $\sum_{n \in \mathbb{N}} a_n e^{ i n \pi x}$.
I am stuck at this point. The definition of a generator requires that we consider the closure of the polynomials of the generator. In this case the terms $\exp( i \pi n x)$ where $n$ is a negative integer are not polynomials.
Am I missing something in the definition of the generator, or am I approaching this problem incorrectly?
Let $\mathbb T_+\subseteq\mathbb C$ be the upper half of the unit circle in the complex plane, i.e. $$\mathbb T=\{z\in\mathbb C \colon |z|=1,\Im z\geq 0\}.$$ Then $C[0,1]$ is isomorphic to $C(\mathbb T_+)$ and the generator $exp(i\pi x)$ corresponds to the function $f(z)=z$. The element $f$ generates the subalgebra of complex polynomials. Since the complement $\mathbb C\setminus \mathbb T_+$ of the spectrum of $C(\mathbb T_+)$ is connected, every continuous function $g:\mathbb T_+\to \mathbb C$ can be uniformly approximated with polynomials, by Mergelyan's theorem.