3

Let $A$ be the algebra of all continuously differentiable complex functions on the interval $[0,1]$ with pointwise multiplication, normed by

$$ ||f||=||f||_{\infty} + ||f'||_{\infty}. $$

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

The problem is finding its maximal ideal space.

As in example 11.13(a) of Rudin's Functional Analysis I try to prove that a maximal ideal is given by the kernel of the complex homomorophism $h_x:A\rightarrow\mathbb{C}$ defined by

$$ h_x(f)=f(x). $$ Then the union of all maximal ideals can be identified with $[0,1]$ and the intersection will be zero function from which it follows that $A$ is semisimple.

Now suppose that a maximal ideal which is not the kernel of any $h_x$ then for every $p\in [0,1]$ there would be a $f \in M$ such that $f(p)\neq 0$. The compactness on $[0.1]$ implies that $M$ contains finitely many functions $f_1, \dots f_n$ such that at least one of them is $\neq 0$ at each point of $[0,1]$. Put

$$ g=f_1\overline{f_1}, \dots f_n\overline{f_n}. $$ If $g\in M$ this would lead to a contradiction since $g$ is invertible and proper ideals don't contain invertible elements.

My question is why is $g\in M$ if it is, implying that it is continuously differentiable on $[0,1]$, or do I have to find a different strategy.

simon
  • 446
  • $g=f_1\overline{f_1}+\dots+f_n\overline{f_n}$ since $M$ is an ideal ($AM\subseteq M$ and $M+M\subseteq M$). – Yurii Savchuk Dec 03 '13 at 08:18
  • Here is where the confusion lies for me. Why is the product of continuously differentiable complex-valued functions is also continuously differentiable. Take $z\in A$ then $\overline{z} \in A$ but the product is not continuously differentiable in $\mathbb{C}$, but is it in $[0,1]$?. – simon Dec 03 '13 at 11:28
  • Your question is basically: why $A$ is an algebra?! In the definition of $A$ you should take real differentiability. Otherwise it is not a Banachalgebra. – Yurii Savchuk Dec 03 '13 at 12:08
  • Indeed it must be that the product of two complex-valued continuously differentiable function on $[0,1]$ is again continuously differentiable in order for $A$ to be an algebra, but why is this true? – simon Dec 03 '13 at 13:52
  • Product of continuously differentiable functions $f,g$ is again differentiable and $(fg)'=f'g+fg'$ is continuous, isn't it? – Yurii Savchuk Dec 03 '13 at 14:07
  • Yes of course, I think I got confused by f being complex valued. – simon Dec 03 '13 at 14:17

0 Answers0