If A is a Banach algebra, then so is A/I

[Fredrik]

Homework Statement

The problem is to prove the following:

If $\mathcal A$ is a Banach algebra, and $\mathcal I$ is a closed ideal in $\mathcal A$, then $\mathcal A/\mathcal I$ is a Banach algebra.

This is problem 3.1.3 (4)(b) in "Functional analysis: spectral theory", by V.S. Sunder. Link.

Homework Equations

A normed algebra is a vector space with a bilinear (i.e. distributive) multiplication operation, and a norm that satisfies the usual conditions on a norm and also $\|xy\|\leq \|x\|\|y\|$. A Banach algebra is a normed algebra in which every Cauchy sequence is convergent.

An ideal I of a normed algebra A is a subspace (in the vector space sense) such that xi=ix is in I for all x in A and all i in I. A closed ideal is an ideal that's also a closed set. In particular, it means that if a sequence in I is convergent, its limit is in I. I'm familiar with a theorem that says that for any x in A, there's a unique i in I such that d(x,I)=d(x,i). This i is such that x-i is orthogonal to I. I suspect that maybe I should use this theorem, because it only holds when I is closed, and nothing I've come up with uses that assumption.

The members of the quotient algebra A/I are subsets of A, which we write as x+I={x+i|i is in I}. We define addition, multiplication by a scalar, multiplication, and the norm by

(x+I)+(y+I)=(x+y)+I
a(x+I)=ax+I
(x+I)(y+I)=xy+I
$\|x+I\|=d(x,I)=\inf\{d(x,i)|i\in I\}$

In part (a), I proved that A/I is a normed algebra.

The Attempt at a Solution

The only strategy I've come up with is the following: Consider a Cauchy sequence $x_n+I$ in A/I and show that $x_n$ must be a Cauchy sequence too. Since A is complete, $x_n\rightarrow x$ for some x. The next step would be to show that $x_n+\mathcal I\rightarrow x+I$

Unfortunately,

$$\|x_n-x_m\|=\|x_n-x_m-0\|\geq d(x_n-x_m,I)=\|(x_n-x_m)+I\|=\|(x_n+I)-(x_m+I)\|$$

so I don't seem to be able to use the assumption that I can make that last thing arbitrarily small. However, if I can show that $x_n\rightarrow x$ for some x, I can handle the rest:

$$\|(x_n+I)-(x+I)\|=\|(x_n-x)+I\|=d(x_n-x,I)\leq d(x_n-x,0)=\|x_n-x\|<\varepsilon$$

 

[micromass]

First I note that you probably had to use closedness of I when you showed that that norm is really a norm on X/I.

To show that it is complete: the trick is to use series and absolute convergence (I assume that you seen that a space is Banach iff every absolute convergent series is convergent, tell me if you need a prood of this).

Take $$\sum{x_n+I}$$ absolute convergent. We will show that the series is convergent. By definition of the quotient norm, there exists a sequence $$z_n\in x_n+I$$ in X, such that

$$\|z_n\|\leq \|(x_n+I\|+2^{-n}$$

Thus the series $$\sum{z_n}$$ converges absolutely. Since absolute convergence implies convergence IN A BANACHSPACE, this implies that there exists a z such that $$z=\sum{z_n}$$. But now we have that

$$z-\sum_{k=1}^n{z_k}+I=(z+I)-\sum_{k=1}^n{x_n+I}$$

Since \|z-\sum_{k=1}^n{z_k}+I\|\rightarrow 0[/tex], this implies that $$\sum_{x_n+I}=z+I$$. Thus our series is absolutely convergent...

 

[Fredrik]

First I note that you probably had to use closedness of I when you showed that that norm is really a norm on X/I.

I actually didn't, but thanks for saying that, because while trying to explain what I did, I found that I had made a really silly mistake. I proved that $\|a(x+I)\|\leq |a|\|x+I\|$ instead of proving equality. I guess I got carried away by the fact that most of the things I've been proving are inequalities. I do have to use that I is closed to prove that $\|a(x+I)\|\geq |a|\|x+I\|$.

To show that it is complete: the trick is to use series and absolute convergence (I assume that you seen that a space is Banach iff every absolute convergent series is convergent, tell me if you need a prood of this).

I'm not familiar with this, but I own three relevant books, Sunder, Conway and Friedman, and one of them must have a proof of that. (I haven't looked yet).

Take $$\sum{x_n+I}$$ absolute convergent. We will show that the series is convergent. By definition of the quotient norm, there exists a sequence $$z_n\in x_n+I$$ in X, such that

$$\|z_n\|\leq \|(x_n+I\|+2^{-n}$$

Thus the series $$\sum{z_n}$$ converges absolutely. Since absolute convergence implies convergence IN A BANACHSPACE, this implies that there exists a z such that $$z=\sum{z_n}$$. But now we have that

$$z-\sum_{k=1}^n{z_k}+I=(z+I)-\sum_{k=1}^n{x_n+I}$$

Since $$\|z-\sum_{k=1}^n{z_k}+I\|\rightarrow 0$$, this implies that $$\sum {x_n+I}=z+I$$. Thus our series is absolutely convergent...

I have worked through the details and understand this proof now. That's very clever. It would have taken me a lot of time to think of that. So thanks again.

I'm still pretty frustrated that I couldn't make my straightforward approach work. Do you think that it can be done that way too, or is it simply not true that if x_n+I is a Cauchy sequence, then x_n is a Cauchy sequence? (That was the step I failed to prove).

 

[micromass]

I don't really think that is true... What I think to be true is, that if $$x_n+I$$ is Cauchy then there exists a sequence of representatives that is a Cauchy sequence. But I don't think that it is necessairy that the xn is a Cauchy sequence.

As for the proof of the thing I left out, i.e. if every absolutely convergent series is convergent, then the space is Banach. Here it is:

Take a Cauchy sequence $$(x_n)_n$$ of X. We will show that this sequence has a convergent subsequence. Since the sequence is Cauchy, there exists a subsequence such that

$$\|x_{k_n}+x_{k_{n-1}}\|\leq 2^{-n}$$

Thus the series $$\sum{x_{k_n}-x_{k_{n-1}}}$$ converges absolutely and thus (by assumption) converges. The sum is a telescopic one, thus we can easily conclude that $$x_{k_n}$$ converges. Thus our Cauchy sequence has a convergent subsequence and is thus convergent.

 

[Fredrik]

I don't really think that is true... What I think to be true is, that if $$x_n+I$$ is Cauchy then there exists a sequence of representatives that is a Cauchy sequence. But I don't think that it is necessairy that the xn is a Cauchy sequence.

Ah...that helps a lot. I was finally able to find a proof that doesn't involve series thanks to this tip.

Let x_n+I be a Cauchy sequence in A/I, and choose N such that

$$n,m\geq N\Rightarrow \|(x_n+I)-(x_m+I)\|<\frac{\varepsilon}{2}$$

If z_n is in x_n+I for each n, we have

$$\|z_n-z_m\|=\|x_n-x_m-(i_n-i_m)\|$$

The i_k with k≥N can be chosen so that for $n,m\geq N$, the above is

$$\leq d(x_n-x_m,I)+\frac{\varepsilon}{2}=\|(x_n-x_m)+I\|+\frac{\varepsilon}{2}<\varepsilon$$

So z_n is a Cauchy sequence, with a limit I'll call x. Now I can prove that $x_n+I\rightarrow x+I$ (which means that an arbitrary Cauchy sequence is convergent).

$$\|(x_n+I)-(x+I)\|=\|(x_n-x)+I\|=\|(z_n-x)+I\|\leq d(z_n-x,0)=\|z_n-x\|<\varepsilon$$.

As for the proof of the thing I left out, i.e. if every absolutely convergent series is convergent, then the space is Banach. Here it is:

Take a Cauchy sequence $$(x_n)_n$$ of X. We will show that this sequence has a convergent subsequence. Since the sequence is Cauchy, there exists a subsequence such that

$$\|x_{k_n}+x_{k_{n-1}}\|\leq 2^{-n}$$

Thus the series $$\sum{x_{k_n}-x_{k_{n-1}}}$$ converges absolutely and thus (by assumption) converges. The sum is a telescopic one, thus we can easily conclude that $$x_{k_n}$$ converges. Thus our Cauchy sequence has a convergent subsequence and is thus convergent.

This one took me some time to get, but I understand it now. Thanks again.

 

[micromass]

Ah...that helps a lot. I was finally able to find a proof that doesn't involve series thanks to this tip.

Let x_n+I be a Cauchy sequence in A/I, and choose N such that

$$n,m\geq N\Rightarrow \|(x_n+I)-(x_m+I)\|<\frac{\varepsilon}{2}$$

If z_n is in x_n+I for each n, we have

$$\|z_n-z_m\|=\|x_n-x_m-(i_n-i_m)\|$$

The i_k with k≥N can be chosen so that for $n,m\geq N$, the above is

$$\leq d(x_n-x_m,I)+\frac{\varepsilon}{2}=\|(x_n-x_m)+I\|+\frac{\varepsilon}{2}<\varepsilon$$

So z_n is a Cauchy sequence, with a limit I'll call x. Now I can prove that $x_n+I\rightarrow x+I$ (which means that an arbitrary Cauchy sequence is convergent).

$$\|(x_n+I)-(x+I)\|=\|(x_n-x)+I\|=\|(z_n-x)+I\|\leq d(z_n-x,0)=\|z_n-x\|<\varepsilon$$.

Hmm, this is a cool proof

 

[Fredrik]

I'm bumping this because I found something weird when I was reviewing the proof that A/I is a normed algebra. To prove that $\|a(x+\mathcal I)\|\geq|a|\,\|x+\mathcal I\|$, this is what I had to do:

Let $i_0$ be the unique member of $\mathcal I$ such that $d(ax,i_0)=d(x,\mathcal I)$.
$$\begin{align*} \|a(x+\mathcal I)\| &=\|ax+\mathcal I\|=d(ax,\mathcal I)=\inf\{d(ax,i)|i\in\mathcal I\}\\ &=d(ax,i_0)=\|ax-i_0\|=|a|\,\Big\|x-\frac{i_0}{a}\Big\|=|a|\,d\Big(x,\frac{i_0}{a}\Big)\\ &\geq|a|\,d(x,\mathcal I)=|a|\,\|x+\mathcal I\| \end{align*}$$

My problem with this is that the proof of the theorem that guarantees the existence of this $i_0$ (theorem 2.5 in Conway) is a theorem for Hilbert spaces. It doesn't use the inner product explicitly, but it uses the parallelogram law: $\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$. So it doesn't appear to be valid for an arbitrary Banach algebra.

So if it's actually true that A/I is a Banach algebra whenever A is a Banach algebra and I is a closed ideal in A, then one of the following must also be true:

1. There's a way to prove the inequality above without reference to i₀.
2. It's always possible to define an inner product on a Banach algebra, which turns it into a Hilbert space.
3. It's possible to prove the theorem that guarantees the existence of i₀, for arbitrary Banach algebras, without using the parallelogram law.

I could use some help figuring out which one of these options is correct.

Edit: I found the answer. It's option 1. It turned out that the calculation that I thought had proved the other inequality was actually a proof of the same inequality. I must have made the calculation and then reversed the direction of the inequality when I copied the result to somewhere else. :grumpy:

This is the other calculation:

Let $i\in\mathcal I$ be arbitrary.
$$\begin{align*} d(ax,i)=\|ax-i\|=|a|\,\Big\|x-\frac{i}{a}\Big\|=|a|\,d\Big(x,\frac{i}{a}\Big)\geq|a|\,d(x,\mathcal I)=|a|\,\|x+\mathcal I\| \end{align*}$$

So $|a|\,\|x+\mathcal I\|$ is a lower bound of $\{d(ax,i)|i\in\mathcal I\}$, and the greatest lower bound of this set is $d(ax,\mathcal I)=\|ax+\mathcal I\|=\|a(x+\mathcal I)\|$. Hence $\|a(x+\mathcal I)\|\geq |a|\,\|x+\mathcal I\|$.

 

[Fredrik]

OK, now I have to bump this for another reason. I discovered that my completeness proof is nonsense, because I chose $z_n$ in a way that made it depend on $\varepsilon$. I wasn't able to fix it, so I looked at your proof again, and now I don't understand it. Why does $\|z_n\|\leq\|x_n+I\|+2^{-n}$ imply that $\sum_n z_n$ is absolutely convergent? I know how to deal with the $2^{-n}$ term of course, but what about the other one?

 

[micromass]

OK, now I have to bump this for another reason. I discovered that my completeness proof is nonsense, because I chose $z_n$ in a way that made it depend on $\varepsilon$. I wasn't able to fix it, so I looked at your proof again, and now I don't understand it. Why does $\|z_n\|\leq\|x_n+I\|+2^{-n}$ imply that $\sum_n z_n$ is absolutely convergent? I know how to deal with the $2^{-n}$ term of course, but what about the other one?

Because the series $$\sum(x_n+I)$$ converges absolutely. This means, of course, that $$\sum\|x_n+I\|$$ converges. Thus

$$\sum\|z_n\|\leq \sum\|x_n+I\|+\sum 2^{-n}$$

And both series converge.

 

[Fredrik]

Thank you. Apparently my problem is that I can't read. I skimmed though the beginning of your proof so fast that I missed what you were actually doing. I somehow thought that the starting assumption was "$x_n+I$ is Cauchy", not "$\sum_n (x_n+I)$ is absolutely convergent".