Inclusion $[0,1]\rightarrow\mathbb{C}$ generates $C^{1}[0,1]$ as a Banach algebra

Question

I am trying to show that the inclusion map $x:[0,1]\rightarrow\mathbb{C}$ generates $A=C^{1}[0,1]$ as a Banach algebra. The first thing that occurred to me was to try using Stone-Weierstrass but somehow I'm not getting it.

Also, I want to show that the map $[0,1]\rightarrow\Omega(A),t\mapsto\tau_{t}$ is onto, where $\Omega(A)$ is the space of characters on $A$ and $\tau_{t}(f)=f(t)$ for $f\in A$. If that is going to be true, then I must have $\phi(x)\in[0,1]$ for all $\phi\in\Omega(A)$ where $x$ is the inclusion map. But I can't see why this is so.

Which version of Stone-Weierstrass are you using? The one on Wikipedia (http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Stone.E2.80.93Weierstrass_theorem.2C_complex_version) directly gives you the result you want. — bradhd, Nov 02 '13 at 22:22
Actually, Weierstrass approximation theorem does the job. Your other question is contained in Theorem 5.1 here. — Julien, Nov 02 '13 at 22:30

score 2 · Accepted Answer · answered Nov 02 '13 at 22:30

For the first it is exactly as you say, the algebra generated by $x$ is the algebra of polynomials on $[0,1]$, and this satisfies the conditions for Stone-Weiertrass.

Regarding your second question, note that $\phi$ is necessarily positive. As the order you will consider is the pointwise one, your have $0\leq x\leq1$, and so $0=\phi(0)\leq\phi(x)\leq\phi(1)=1$.

Inclusion $[0,1]\rightarrow\mathbb{C}$ generates $C^{1}[0,1]$ as a Banach algebra

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