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Is $M_a=\{f\in \mathcal C[0,1]:f(a)=0\}$ where $\mathcal C[0,1]$ denotes the ring of continuous functions in $[0,1]$ a Principal ideal?

My try: Let $M_a=\langle f_1\rangle$ .Consider $f(x)=\sqrt {|f_1(x)|}$ then $f(x)=g_1(x)f_1(x)$ .

Now $|g_1(x)|\le M$ since image of a continuous function on a compact set is bounded.

Also by continuity of $f_1(x)$ at $a$ if we choose $\epsilon=\frac{1}{M^2}$ then there exists $U$ an open set containing $a$ such that $|f_1(x)|\le \frac{1}{M^2}$ $\forall x\in U$.

Now for all $x\in U$ , $|f(x)|\le M|f_1(x)|\le M\frac{1}{M^2}=\frac{1}{M}$

But I am not getting any contradiction from here.Please suggest some steps

Learnmore
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    Dear @learnmore Infact, $M_a$ is not even countably generated.See here:http://math.stackexchange.com/questions/1447559/m-a-f-in-c0-1-fa-0-for-a-in-0-1-is-m-a-finitely-gene Regards, – Arpit Kansal May 07 '16 at 04:27
  • @ArpitKansal;I saw it however I could not understand the inequality $M\sum |f_k(x)|<\sum \sqrt{|f_k(x)}$ there can you explain that to me – Learnmore May 07 '16 at 05:29
  • Dear @learnmore,just use that if $x >0$ then $ x= \sqrt x \sqrt x $ and geometric series .Regards, – Arpit Kansal May 07 '16 at 10:41
  • I did not get that @ArpitKansal; $M\sum |f_k(x)|=M\sum \sqrt {|f_k(x)}\sqrt {|f_k(x)|}$ – Learnmore May 07 '16 at 12:23
  • How to get the inequality from here@ArpitKansal – Learnmore May 07 '16 at 12:23
  • Dear @learnmore,But you already know that: $$\sqrt{|f_i(x) |} \le \frac{1}{2^{i+1}M} \ \ \ \ \ \forall x\in U, i=1, \cdots r.$$ – Arpit Kansal May 07 '16 at 12:33

3 Answers3

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The idea is right, and little modification is needed.

For any given $\epsilon\gt 0$, there is a neighbourhood of $a$ such that if $x\ne a$ is in that neighbourhood, then $0\ne |f_1(x)|\lt \epsilon$.

Let $x\ne a$ be in that neighbourhood, and suppose that $|f_1(x)|=\theta$. Then $\sqrt{|f_1(x)|}=\sqrt{\theta}$ and therefore $g_1(x)=\frac{1}{\sqrt{\theta}}$. Thus $g_1$ is unbounded.

André Nicolas
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It is not principal.

Suppose $M_a$ is generated by $f_1$, then $f_1$ has only one zero, $a$ since if it has another zero $b\in [a,b]$ there exists $g:[a,b]\rightarrow R$ continuous such that $g(b)\neq 0$ and $g(a)=0$. Without restricting the generality, we can suppose:

$f_1$ has a constant sign, in this case I adapt your proof, by without restricting the generality, supposing that $f_1\geq 0$, let $g=\sqrt f_1$, $g=hf_1$. This implies that for $x\neq a$, $h={g\over {f_1}}={1\over {\sqrt f_1}}$, but ${1\over{\sqrt f_1}}$ can't be extended by continuity at zero. done.

Suppose that $f_1$ does not have a constant sign, without restricting the generality, we can suppose that $f_1(x)<0, x<a$ and $f_1(x)>0, x>a$ consider $g=\mid f_1\mid$, it is a continuous function and $g=hf_1$, this implies that $h=-1$ if $x<a$ and $h=1, x>a$ but $h$ is not continuous. done.

Tsemo Aristide
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Suppose $M_a$ is generated by $f\in C[0,1].$

(1). Show that $f(x)\ne 0$ for $x\ne a.$

(2). Show that if $\sqrt {|f(x)|}=f(x)g(x)$ for all $x$, for some $g\in C[0,1],$ then,using (1), that $g(x)$ is not continuous at $x=a.$

DanielWainfleet
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