Show that $I_A=\{f\in C[0,1]:\forall x\in A ,f(x)=0\}$ is closed ideal of $C[0,1]$

Question

let $A$ be closed subset of $[0,1]$, then $$I_A=\{f\in C[0,1]:\forall x\in A ,f(x)=0\}$$ is closed ideal of $C[0,1]$ what are the maximal ideals??

i have proved that $I_A$ is ideal of Banach algebra $C[0,1]$.

but how to show that it is closed ideal. i have approach it like this.

let $(f_n)$ be any sequence in $I_A$ such that it converge to $f$ in $C[0,1]$ .since convergence in $C[0,1]$ is uniform, in particular pointwise. then $f(x)=0$ for all $x\in A $ since $f_n(x)=0$

hence $f\in I_A$. Is this right method??

what about maximal ideals? any hint.

thanks in advanced.

Answer 1

Your approach is correct. Another possibility consists in proving that $I_A$ is closed by proving that its complement is open. If $f\notin I_A$, then $f(a)\neq0$ for some $a\in A$. Take $r=\bigl|f(a)\bigr|$. Then$$B(f,r)\subset C\bigl([0,1]\bigr)\setminus I_A,$$because if $g\in B(f,r)$, then$$\bigl|g(a)\bigr|\geqslant\bigl|g(a)\bigr|-\bigl|g(a)-f(a)\bigr|>r-d(f,g)>0.$$

Answer 2

Note that for $B \subset A$ we always have $I_A \subset I_B$. Thus for $x \in A$, we know that $I_A \subset I_{\{x\}}$. Both are ideals. What happens if $A \neq \{x\}$? (In this case $I_A$ cannot be maximal.)

Now suppose that $I_{\{x\}} \subset I$ for an proper ideal $I$. Then there must exist $g \in I$ with $g(x) \neq 0$. By multypling with a bump-function, we can suppose that $g$ is supported in a neighborhood of $x$ and $g(x)=1$ by rescaling with a scalar.

For any $f \in C([0,1])$, we have $fg \in I$ and $f(1-g) \in I_x$ because $f(x)(g(x)-1))=0$. Thus $f= fg+f(1-g) \in I$. Thus $I = C([0,1])$.

Update: I have only proved that $I_x$ is a maximal ideal. Let us prove that any maximal ideal is already given by some $I_x$. Assume that $I$ is a maximal ideal with $I \neq I_x$ for all $x \in [0,1]$. Since $I_x$ is maximal, this implies that $I_x \not\subset I$, i.e. for any $x \in [0,1]$ there exists $f \in I$ with $f_x(x) \neq 0$.

By multiplying with a bump-function and, if necessary, with a constant we can construct $g_x \in I$ such that $g_x$ is non-negative and $g_x(x) =1$. Let $U_x$ be a neighborhood of $x$ with $g_x(y) \neq 0$ for all $y \in U_x$. Since $[0,1]$ is compact, we can cover it with finite many $U_{x_1},\ldots,U_{x_n}$. The function $g:= g_{x_1}+\ldots+g_{x_n}$ is in $I$ and has no zeros, i.e. is bounded below. This shows that $1= \frac{1}{g} g \in I$. Hence $I = C([0,1])$.

Answer 3

To find the maximal ideals you first want to prove the converse:

Proposition If $I$ is a closed ideal in $C([0,1])$ there exists $A$ so that $I=I_A$.

That takes a little work.

Lemma 0. If $I$ is an ideal and there exists $f\in I$ such that $f(x)\ne 0$ for all $x\in[0,1]$ then $I=C([0,1])$.

Hint: The definition of "ideal"...

Lemma 1. Suppose $I$ is an ideal and for every $x\in[0,1]$ there exists $f\in I$ with $f(x)\ne0$. Then $I=C([0,1])$.

Deriving this from Lemma 0 is the main trick. Hint: If $f\in I$ then $|f|^2\in I$, because $|f|^2=f\overline f$.

Now assume I is a proper closed ideal. Let $A$ be the intersection of the zero set of $f$ for all $f\in I$. Then $I\subset I_A$, and with a little work you can prove the opposite inclusion. (Lemma 1 says precisely that $A\ne\emptyset$. Edit: Come to think of it, if you just want to find the maximal ideals you don't need the other half; it's enough to show that if $I$ is a proper ideal there exists $A\ne\emptyset$ with $I\subset I_A$.)

Hints on how to show $I_A\subset I$: First show that if $f$ vanishes on a neighborhood of $A$ then $f\in I$. Then show that if $f\in I_A$ there exists a sequence $(f_n)$ such that each $f_n$ vanishes on a neighborhood of $A$ and $f_n\to f$ uniformly.

Given the proposition it's easy to find the maximal ideals: $I_A\subset I_B$ if and only if $A$ and $B$ satisfy _what_condition?