If $X$ is a Banach algebra with identity, and $0$ is the only element $x \in X$ such that there is a sequence $\{ {x_n}\} \subset X$, $\left\| {{x_n}} \right\| = 1$ and $x{x_n} \to 0$ or ${x_n}x \to 0$, then is it necessarily that every non-zero element is invertible?
a question about invertibility of Banach Algebra
-
0It is possible there is a $x\in X$ s.t $x$ is not invertible, because as $n\to\infty$, we have $1=\|x_{n}\|=\|(1-x)x_{n}\|\leq\|(1-x)\|\|x_{n}\|=\|1-x\|$ so, $\|1-x\|\geq1$ we can assume that for some $x\in X$, $\|1-x\|>1$.(we know that if $\|1-x\|<1$ then $x$ is invertible.. – 2011-12-22
-
0@Vahid: That doesn't really help. If $\|1-x\|<1$, then $x$ is invertible, but the converse doesn't hold. In any case, it is clear that such $x$ as described in the hypothesis, even if it isn't assumed to be $0$, is not invertible. The question is whether $0$ being the only such element implies that every nonzero element is invertible. – 2011-12-23
1 Answers
The answer is yes. We have $X=\mathbb Ce$, which is in fact equivalent to the fact that every non-zero element is invertible by Gelfand-Mazur theorem.
We have the following results:
Lemma 1. Let $x\in X$ and $\{x_n\}$ a sequence in $X$, which converges to $x$ and such that we can find a bounded sequence $\{y_n\}$ such that $x_ny_n=e$. Then $x$ is (left)-invertible.
Proof: Let $M:=\sup_{k\in\mathbb N} \|y_k \|<\infty$. Let $n_0$ such that $\|x-x_{n_0}\|\leq \frac 1{2M}$. Then $\|x-x_{n_0}\|< \|y_{n_0}\|^{-1}$. Then $(x-x_{n_0})y_{n_0}$ is invertible, and $x=((x-x_{n_0})y_{n_0}+e)x_{n_0}$, so $xy_{n_0}=(x-x_{n_0})y_{n_0}+e$ and $xy_{n_0}\sum_{k=0}^{+\infty}(-1)^k((x-x_{n_0})y_{n_0})^k=e$.
Definition. If $x$ satisfies "we can find $\{x_n\}\subset X$ such that $\|x_n\|=1$ and $x_nx\to0$ (respectively $xx_n\to 0$), then $x$ is a right (respectively left) topological divisor of $0$.
Lemma 2. An element of the boundary of the set of invertible elements of $X$, $\mathcal I$, is a left and right topological divisor of $0$.
Proof: Let $x\in\partial \mathcal I$. Since $\mathcal I$ is open, $x$ is not invertible and we can find a sequence of right invertible elements (for left invertible it's the same proof) $\{z_n\}$ such that $z_n\to x$. Let $y_n$ such that $z_ny_n=e$. Since $x$ is not invertible, by lemma 1. we can assume, after taking a subsequence, that $\|y_n\|\geq n$. Put $x_n:=\frac 1{\|y_n\|}y_n$. Then $$\|xx_n\| = \|(x-z_n)x_n\|+\|z_nx_n\|\leq \|x-z_n\|+\frac{\|e\|}{\|y_n\|}\leq \|x-z_n\|+\frac 1n.$$
Now we can show that $X=\mathbb C e$. (the initial proof was not complete, but thanks to @Jonas Meyer it is, see comments below) Indeed, let $x\in X$, and $\lambda_0$ in the boundary of the spectrum of $x$, $\operatorname{Sp}(x)$. Since $\operatorname{Sp}(x)$ is closed, $x-\lambda_0 e$ is not invertible, and we can find invertible elements of the form $x-\lambda_n e$ such that $x-\lambda_n e\to x-\lambda_0 e$. So by lemma 2., we have $x-\lambda_0e=0$ and $x\in\mathbb C e$.
So $x=\lambda_0e$.
-
1"...which is stronger": By the Mazur-Gelfand theorem, it is equivalent. "We can find a sequence $\{\lambda_n\}$ contained in the interior": I don't understand why you are assuming that the spectrum of $x$ has nonempty interior. "either $\lambda_0e-x\in\mathcal{I}$..." Since $\lambda_0$ is in the boundary of the spectrum and the spectrum is closed, this is clearly impossible. Also, $X\setminus \mathcal I$ is closed, so $\partial\mathcal I=\partial(X\setminus\mathcal I)\subseteq (X\setminus\mathcal I)$. – 2011-12-23
-
1But, I think I see the point. Given $x$, take $\lambda_0$ in the boundary of the spectrum of $x$, and then $\lambda_0e-x$ is in $\partial I$, and then $\lambda_0 e-x$ is a left and right topological divisor of $0$ by Lemma 2, and then $x=\lambda_0 e$ by hypothesis. +1 – 2011-12-23
-
0@JonasMeyer You are right, I didn't realize that it's equivalent and I will correct it. – 2011-12-23
-
0Davide, Thank you. However, while I like this answer, there are still 2 things that I think could be cleaned up. (1) There is no reason to assume that the spectrum $x$ has nonempty interior, and even if the interior is nonempty, the existence of such $\{\lambda_n\}$ doesn't follow. Also, the stuff about the interior is superfluous, even if such $\{\lambda_n\}$ does exist. (2) There is no reason to include an additional argument for why $\lambda_0 e-x$ is not invertible. This confuses me in part because you noticed this when writing the proof of lemma 2. $\mathcal I$ is open. – 2011-12-23
-
0All that is needed once you have the good idea to consider $\lambda_0$ in the boundary of the spectrum is that therefore (a) $x-\lambda_0 e$ is not invertible (the spectrum is closed), (b) $x-\lambda_0 e$ is arbitrarily close to invertible elements (of the form $x-\lambda e$). Combining (a) and (b), $x-\lambda_0 e\in\partial I$, and then direct application of lemma 2 and the hypothesis gives $x-\lambda_0 e=0$. – 2011-12-23
-
0@JonasMeyer: yes indeed, the spectrum of $x$ can have an empty interior. I will put your argument. – 2011-12-23
-
0Thank you. Now it occurs to me, this answer is nice in part because it shows directly that $X\cong\mathbb C$ without having to invoke the Mazur-Gelfand theorem. One could give an argument without using the spectrum, but it wouldn't have this feature. – 2011-12-23
-
1@JonasMeyer : Actually, the Gelfand-Mazur theorem is a direct consequence of the fact that the spectrum is always non-empty, and the answer uses that fact. – 2011-12-23
-
0@MalikYounsi: Thanks, good point. – 2011-12-24