Maximal ideals in the algebra of continuously differentiable functions on [0,1] This is an exercise in Rudin's Functional Analysis, in the chapter on commutative Banach algebras. My (uneducated) guess was that every homomorphism on $C^{1}[0,1]$ is an evaluation at some point of [0,1]. The imitation of the proof for finding the homomorphisms on $C[0,1]$, as in Simmons, fails as taking the moduli of functions does not preserve differentiability. I would be grateful for hints on this.
 A: Yes, every (unital) homomorphism on $C^1([0,1])$ is an evaluation homomorphism.
Proof 1: Let $x\in C^1([0,1])$ the identity, i.e. $x(t) = t$ for all $t\in [0,1]$. Let $\varphi \colon C^1([0,1]) \to \mathbb{C}$ a (unital) algebra homomorphism, and $z := \varphi(x)$. Then $z \in [0,1]$ - otherwise $x - z$ would be invertible - and hence we have
$$\varphi(P) = P(z)$$
for all polynomials $P$. We will show below that the polynomials form a dense subspace of $C^1([0,1])$, then $\varphi(f) = f(z)$ for all $f\in C^1([0,1])$ follows by continuity.
To see that the subspace of polynomials is dense, let $f\in C^1([0,1])$ and $\varepsilon > 0$ arbitrary. By the Stone-Weierstraß theorem, there is a polynomial $P_1$ with $\lVert f' - P_1\rVert_\infty < \varepsilon/2$. Let
$$P(t) = f(0) + \int_0^t P_1(s)\,ds.$$
Then $\lVert f - P\rVert < \varepsilon$:
$$\begin{align}
\lvert f(t) - P(t)\rvert &= \left\lvert f(0) + \int_0^t f'(s)\,ds - \left(f(0) + \int_0^t P_1(s)\,ds\right) \right\rvert\\
&= \left\lvert \int_0^t (f'(s) - P_1(s))\,ds\right\rvert\\
&\leqslant \int_0^t \lvert f'(s) - P_1(s)\rvert\,ds\\
&\leqslant t\cdot \lVert f' - P_1\rVert_\infty\\
&\leqslant \frac{\varepsilon}{2},
\end{align}$$
whence $\lVert f - P\rVert = \lVert f-P\rVert_\infty + \lVert f' - P_1\rVert_\infty < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.

Proof 2: Let $\mathfrak{a} \subsetneqq C^1([0,1])$ an ideal. Then
$$V(\mathfrak{a}) = \left\lbrace t \in [0,1] : \bigl(\forall f\in \mathfrak{a}\bigr)\bigl(f(t) = 0\bigr)\right\rbrace \neq \varnothing.$$
Suppose $V(\mathfrak{a}) = \varnothing$. Then, for every $t\in [0,1]$, there would be an $f_t \in \mathfrak{a}$ with $f_t(t) \neq 0$. Since $\mathfrak{a}$ is an ideal, $g_t \colon s \mapsto f_t(s)\cdot \overline{f_t(s)}$ also belongs to $\mathfrak{a}$, and there is an open neighbourhood $U_t$ of $t$ with $g_t(s) > 0$ for all $s \in U_t$. Since $[0,1]$ is compact, there are finitely many points $t_1,\dotsc,t_n$ such that
$$[0,1] = \bigcup_{k=1}^n U_{t_k}.$$
Since all $g_t$ are non-negative, and strictly positive on $U_t$,
$$g = \sum_{k=1}^n g_{t_k}$$
is strictly positive on all of $[0,1]$. But that means $g$ is a unit in $C^1([0,1])$, and hence $\mathfrak{a} = C^1([0,1])$, contradicting the assumption.
Let now $\varphi$ a (unital) homomorphism, and $\ker \varphi = \mathfrak{m} \subsetneqq C^1([0,1])$ the corresponding maximal ideal. Let $p \in V(\mathfrak{m})$. Then $\mathfrak{I}(p) = \left\lbrace f\in C^1([0,1]) : f(p) = 0\right\rbrace$ is a proper ideal containing $\mathfrak{m}$, and by maximality, $\mathfrak{m} = \mathfrak{I}(p) = \ker \eta_p$, where $\eta_p$ is the evaluation homomorphism $f \mapsto f(p)$. Hence $\varphi = \eta_p$.
