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?

  • $\begingroup$ It 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.. $\endgroup$
    – Vahid
    Dec 22 '11 at 11:34
  • $\begingroup$ @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. $\endgroup$ Dec 23 '11 at 4:33

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 $x=\lambda_0e$.

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    $\begingroup$ "...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 $\partial\mathcal I=\partial(X\setminus\mathcal I)\subseteq (X\setminus\mathcal I)$. $\endgroup$ Dec 23 '11 at 4:16
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    $\begingroup$ But, 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 $\endgroup$ Dec 23 '11 at 4:19
  • $\begingroup$ @JonasMeyer You are right, I didn't realize that it's equivalent and I will correct it. $\endgroup$ Dec 23 '11 at 9:26
  • $\begingroup$ Davide, 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. $\endgroup$ Dec 23 '11 at 16:39
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    $\begingroup$ @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. $\endgroup$ Dec 23 '11 at 21:03

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