# Commutative Banach algebra and its maximal ideal space

Let $A:=C^{(n)}([0,1])$ be the set consisting of the n-times continuously differentiable complex-valued functions. Consider $A$ with the norm $$\|f\|:=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}{\dfrac{|f^{(k)}(t)|}{k!}}.$$

I want to show that $A$ is a commutative Banach algebra and find its maximal ideal space.

I know that the following holds:

• for all $f,g\in A$: $(fg)(x)=f(x)g(x)=g(x)f(x)=(gf)(x)$;
• for all $f,g,h\in A$: $((fg)h)(x)=(fg)(x)h(x)=g(x)f(x)h(x)=f(x)(gh)(x)=(f(gh))(x)$;
• for all $f,g,h\in A$: $(f(g+h))(x)=f(x)(g+h)(x)=f(x)(g(x)+h(x))=f(x)g(x)+f(x)h(x)=(fg)(x)+ (fh)(x)=(fg+fh)(x)$;
• hence $(g+h)f=gf+hf$;
• for all $f,g\in A,\alpha\in \mathbb{R}$: $(\alpha(fg))(x)=\alpha(fg)(x)=\alpha f(x)g(x)=(\alpha f)(x)g(x)=((\alpha f)g)(x)= f(x) \alpha g(x)=(f(\alpha g))(x)$;

Now,let $f,g\in A$, then we see that \begin{align} \|fg\|&=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}{\dfrac{|(fg)^{(k)}(t)|}{k!}}\\ &=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}{\dfrac{|\sum_{j=0}^k {k \choose j} \cdot f^{(k-j)}(t)\cdot g^{(j)}(t)|}{k!}}\\ &=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}{\dfrac{|\sum_{j=0}^k {\dfrac{k!}{j!(k-j)!}} \cdot f^{(k-j)}(t)\cdot g^{(j)}(t)|}{k!}}\\ &\leq \max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{k-j}(t)g^k(t)|}{j!(k-j)!}}\\ &=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{k-j}(t)}{(k-j)!}\dfrac{g^k(t)|}{j!}}\\ &=? \end{align}

Obviously, I want to show that $\|fg\|\leq\|f\|\|g\|$, but I dont know how to do this. Any hints?

Next, I want to find its maximal ideal space. I know that $f\equiv 1$ implies that $\|f\|=1$, so $A$ is unital. Hence, we working in a commutative unitial Banach algbra. Therefore, the kernel of a multiplicative linear functional on $A$ is a maximal ideal in $A$. However, I can't find this kernel explicitly

Any hints are appreciated.

## 1 Answer

A typo slipped in; a $k$ became $j$ for no reason. Fixing that, you're almost there, re showing it's a Banach algebra:

\begin{align}\dots=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{(k-j)}(t)}{(k-j)!}\dfrac{g^{(j)}(t)|}{j!}} &=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=j}^{n}\dfrac{|f^{(k-j)}(t)|}{(k-j)!} \\&=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=0}^{n-j}\dfrac{|f^{(k)}(t)|}{k!} \\&\le \max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=0}^{n}\dfrac{|f^{(k)}(t)|}{k!} \\&=||g||\,||f||. \end{align}

The maximal ideal space is $[0,1]$ (that is, every complex homomorphism is evaluation at some point). You need to show that if $I$ is an ideal and for every $t\in[0,1]$ there exists $f\in A$ with $f(t)\ne 0$ then $I=A$. Look up the proof of the corresponding fact for $C([0,1])$; exactly the same argument works here.