= Solution
Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The <product rule> gives
$$
\|fg\|_\infty+\|(fg)'\|_\infty
\leq\|f\|_\infty\|g\|_\infty+\|f'\|_\infty\|g\|_\infty+\|f\|_\infty\|g'\|_\infty
\leq\|f\|\|g\|.
$$
To prove completeness, let $(f_n)$ be a <Cauchy sequence> in this norm. The functions and their derivatives converge uniformly to <continuous functions> $f,g$. Passing to the limit in $f_n(t)-f_n(0)=\int_0^tf_n'(s)\,ds$ gives $f(t)-f(0)=\int_0^tg(s)\,ds$. The <fundamental theorem of calculus> yields $f'=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently \b[$A$ is a unital commutative <Banach algebra>].
Let $u(t)=t$. If $\lambda\notin[0,1]$, then $(u-\lambda1)^{-1}(t)=1/(t-\lambda)$ belongs to $C^1[0,1]$. If $\lambda\in[0,1]$, the function $u-\lambda1$ vanishes somewhere and cannot have a pointwise inverse. Thus $\sigma_A(u)=[0,1]$. Every <character of an algebra> satisfies $\phi(u)=t_0$ for some $t_0\in[0,1]$, by part (d), and therefore $\phi(p(u))=p(t_0)$ for every <polynomial> $p$.
<Polynomials> are dense in this stronger norm, not just the <supremum norm>. By the <Weierstrass approximation theorem>, approximate the real and imaginary parts of $f'$ uniformly by <polynomials>, combining them into $q_n\to f'$. Put $p_n(t)=f(0)+\int_0^tq_n(s)\,ds$. Then $p_n$ is a <polynomial> and
$$
\|p_n'-f'\|_\infty\to0,\qquad\|p_n-f\|_\infty\leq\|q_n-f'\|_\infty\to0.
$$
<Continuity> of the <algebra character> now gives $\phi(f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous <algebra character>. Hence the <character space of C1 on a compact interval> consists exactly of the evaluation functionals, and part (c) gives
$$
\boxed{\mathcal M_A=\{M_t:t\in[0,1]\},\qquad M_t=\{f\in C^1[0,1]:f(t)=0\}.}
$$
Distinct points give distinct <maximal ideals>, since the coordinate function separates them.
Back to article page
