# Left topological zero-divisors in Banach algebras.

Let $A$ be a unital Banach algebra. Define $\zeta: A \longrightarrow [0,\infty)$ by $$\forall a \in A: \quad \zeta(a) \stackrel{\text{def}}{=} \inf_{b \in \mathbb{S}(A)} \| ab \|,$$ where $\mathbb{S}(A)$ denotes the unit sphere of $A$.

Definition An element $a \in A$ is called a left topological zero-divisor iff $\zeta(a) = 0$, or equivalently, iff there exists a sequence $(b_{n})_{n \in \mathbb{N}}$ in $\mathbb{S}(A)$ such that $\displaystyle \lim_{n \to \infty} a b_{n} = 0$.

(a) Prove that a left topological zero-divisor is not invertible.

(b) Prove that $\zeta: A \longrightarrow [0,\infty)$ is continuous.

(c) Let $a \in \partial(G(A))$, the boundary of $G(A)$. Prove that there exists a sequence $(\nu_{n})_{n \in \mathbb{N}}$ of invertible elements in $A$ such that $\displaystyle \lim_{n \to \infty} \nu_{n} = a$ and $\displaystyle \lim_{n \to \infty} \| \nu_{n}^{-1} \| = \infty$.

Note: $G(A) \stackrel{\text{def}}{=} \{ a \in A \mid \text{$ a $is invertible} \}$.

• What have you done so far? Where are you stuck? – Robert Israel Jan 31 '13 at 0:58
• @RobertIsrael I only tried to do part (a), I tried to do it with contradiction, but I can't get there for some reason. – i.a.m Jan 31 '13 at 1:27
• @RobertIsrael thank you for the hint, so we have the following, we have,if $a$ is invertible then $\|a\|\le \frac{1}{\|a^{-1}\|}=\frac{\|b_n\|}{\|a^{-1}\|}\le \frac{\|a^{-1}\| \|ab_n\|}{\|a^{-1}\|}=\|ab_{n}\|\longrightarrow0.$ which imples that $\|a\|=0$ and hence $a=0$ contradiction. – i.a.m Jan 31 '13 at 4:21
• @RobertIsrael for b) let $a_n\longrightarrow a$ then we have $\zeta(a_n)=\inf\|a_nb\|=\inf\|ab\|=\zeta(a)$. is this true, can I take the limit inside the inf because the inf has nothing to do with n? – i.a.m Jan 31 '13 at 4:41
• No, $\|a\| \ge 1/\|a^{-1}\|$, not $\le$. Try $\|b_n\| \le \|a^{-1}\| \|a b_n\|$. – Robert Israel Jan 31 '13 at 7:50

In what follows, $\mathcal{A}$ shall denote a unital Banach algebra and $\mathbb{S}(\mathcal{A})$ the unit sphere of $\mathcal{A}$.

(a) Let $a$ be an invertible element of $\mathcal{A}$. Assume, for the sake of contradiction, that there exists a sequence $(b_{n})_{n \in \mathbb{N}}$ in $\mathbb{S}(\mathcal{A})$ such that $\displaystyle \lim_{n \to \infty} a b_{n} = \mathbf{0}_{\mathcal{A}}$. As left-multiplication by any element of $\mathcal{A}$ is a continuous operation on $\mathcal{A}$, we obtain \begin{align} \lim_{n \to \infty} b_{n} &= \lim_{n \to \infty} a^{-1} (a b_{n}) \\ &= a^{-1} \left( \lim_{n \to \infty} a b_{n} \right) \\ &= a^{-1} \cdot \mathbf{0}_{\mathcal{A}} \\ &= \mathbf{0}_{\mathcal{A}}. \end{align} This clearly contradicts the requirement that $\forall n \in \mathbb{N}: ~ b_{n} \in \mathbb{S}(\mathcal{A})$, so it must be the case that $$\{ ab \in \mathcal{A} ~|~ b \in \mathbb{S}(\mathcal{A}) \}$$ is bounded away from $\mathbf{0}_{\mathcal{A}}$, which yields $\zeta(a) > 0$. By taking the contrapositive of this conclusion, Problem (a) is thereby solved.

(b) Fix $a \in \mathcal{A}$, and let $(a_{n})_{n \in \mathbb{N}}$ be a sequence in $\mathcal{A}$ that converges to $a$.

Claim 1: $\displaystyle \limsup_{n \to \infty} \zeta(a_{n}) \leq \zeta(a)$.

Proof of Claim 1: Let $\epsilon > 0$, and find a $b \in \mathbb{S}(\mathcal{A})$ such that $$\zeta(a) \leq \| ab \| < \zeta(a) + \epsilon.$$ Next, observe that

• $\forall n \in \mathbb{N}: ~ \zeta(a_{n}) \leq \| a_{n} b \|$ and

• $\displaystyle \lim_{n \to \infty} \| a_{n} b \| = \| ab \|$.

Hence, $\zeta(a_{n}) < \zeta(a) + \epsilon$ for all $n \in \mathbb{N}$ sufficiently large, which yields $\displaystyle \limsup_{n \to \infty} \zeta(a_{n}) \leq \zeta(a) + \epsilon$. As $\epsilon$ is arbitrary, we obtain $\displaystyle \limsup_{n \to \infty} \zeta(a_{n}) \leq \zeta(a)$. $\quad \spadesuit$

Claim 2: $\displaystyle \zeta(a) \leq \liminf_{n \to \infty} \zeta(a_{n})$.

Proof of Claim 2: Let $\epsilon > 0$, and pick a sequence $(b_{n})_{n \in \mathbb{N}}$ in $\mathbb{S}(\mathcal{A})$ such that $$\forall n \in \mathbb{N}: \quad \zeta(a_{n}) \leq \| a_{n} b_{n} \| < \zeta(a_{n}) + \epsilon.$$ Next, observe that \begin{align} \forall n \in \mathbb{N}: \quad |\| a_{n} b_{n} \| - \| a b_{n} \|| &\leq \| a_{n} b_{n} - a b_{n} \| \\ &= \| (a_{n} - a) b_{n} \| \\ &\leq \| a_{n} - a \| \| b_{n} \| \\ &= \| a_{n} - a \|. \quad (\text{As $\| b_{n} \| = 1$ for all $n \in \mathbb{N}$.}) \end{align} Hence, $\displaystyle \lim_{n \to \infty} (\| a_{n} b_{n} \| - \| a b_{n} \|) = 0$, from which it follows that $$\zeta(a) \leq \| a b_{n} \| < \zeta(a_{n}) + 2 \epsilon$$ for all $n \in \mathbb{N}$ sufficiently large. This yields $\displaystyle \zeta(a) - 2 \epsilon \leq \liminf_{n \to \infty} \zeta(a_{n})$, and as $\epsilon$ is arbitrary, we obtain $\displaystyle \zeta(a) \leq \liminf_{n \to \infty} \zeta(a_{n})$. $\quad \spadesuit$

By the two claims, $\displaystyle \lim_{n \to \infty} \zeta(a_{n}) = \zeta(a)$. Therefore, as $a$ is arbitrary, we conclude that $\zeta$ is a continuous function.

(c) Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence in $\mathcal{G}(\mathcal{A})$ that converges to some $a \in \mathcal{A}$. We claim that if $(\| a_{n}^{-1} \|)_{n \in \mathbb{N}}$ is a bounded sequence in $\mathbb{R}_{+}$, then $a \in \mathcal{G}(\mathcal{A})$. Indeed, suppose that $(\| a_{n}^{-1} \|)_{n \in \mathbb{N}}$ is bounded above by $M > 0$. Then \begin{align} \forall m,n \in \mathbb{N}: \quad \| a_{m}^{-1} - a_{n}^{-1} \| &= \| a_{m}^{-1} a_{n}^{-1} (a_{n} - a_{m}) \| \\ &\leq \| a_{m}^{-1} \| \| a_{n}^{-1} \| \| a_{n} - a_{m} \| \\ &\leq M^{2} \| a_{n} - a_{m} \|. \end{align} As $(a_{n})_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathcal{A}$, it follows that $(a_{n}^{-1})_{n \in \mathbb{N}}$ is also a Cauchy sequence in $\mathcal{A}$. By the completeness of $\mathcal{A}$, we see that $\displaystyle \lim_{n \to \infty} a_{n}^{-1} = b$ for some $b \in \mathcal{A}$. Then as multiplication in $\mathcal{A}$ is a jointly continuous binary operation on $\mathcal{A}$, we obtain \begin{align} ba = \lim_{n \to \infty} a_{n}^{-1} a_{n} = \mathbf{1}_{\mathcal{A}}, \\ ab = \lim_{n \to \infty} a_{n} a_{n}^{-1} = \mathbf{1}_{\mathcal{A}}. \end{align} Therefore, $a$ is invertible and $\displaystyle \lim_{n \to \infty} a_{n}^{-1} = a^{-1}$.

Now, as $\mathcal{G}(\mathcal{A})$ is known to be an open subset of $\mathcal{A}$, we have $\partial(\mathcal{G}(\mathcal{A})) = \text{cl}(\mathcal{G}(\mathcal{A})) \setminus \mathcal{G}(\mathcal{A})$. Let $a \in \partial(\mathcal{G}(\mathcal{A}))$. Then

• $a \notin \mathcal{G}(\mathcal{A})$ and

• there exists a sequence $(a_{n})_{n \in \mathbb{N}}$ in $\mathcal{G}(\mathcal{A})$ that converges to $a$.

By the previous paragraph, $(\| a_{n}^{-1} \|)_{n \in \mathbb{N}}$ is necessarily an unbounded sequence in $\mathbb{R}_{+}$.

Question raised by the OP in his comment below: Is every $a \in \partial(\mathcal{G}(\mathcal{A}))$ a left topological zero-divisor?

The answer is ‘yes’. Fix $a \in \partial(\mathcal{G}(\mathcal{A}))$, and let $(a_{n})_{n \in \mathbb{N}}$ be a sequence in $\mathcal{G}(\mathcal{A})$ converging to $a$ such that $\displaystyle \lim_{n \to \infty} \| a_{n}^{-1} \| = \infty$. Then for all $n \in \mathbb{N}$, we have \begin{align} \left\| a \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \right\| &\leq \left\| a \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} - a_{n} \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \right\| + \left\| a_{n} \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \right\| \quad (\text{By the Triangle Inequality.}) \\ &= \left\| (a - a_{n}) \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \right\| + \left\| \frac{\mathbf{1}_{\mathcal{A}}}{\| a_{n}^{-1} \|} \right\| \\ &\leq \| a - a_{n} \| \cdot \left\| \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \right\| + \left\| \frac{\mathbf{1}_{\mathcal{A}}}{\| a_{n}^{-1} \|} \right\| \\ &= \| a - a_{n} \| + \frac{1}{\| a_{n}^{-1} \|}. \end{align} As the last line converges to $0$ as $n \to \infty$, we obtain $\displaystyle \lim_{n \to \infty} a \cdot \frac{a_{n}^{-1}}{\| a_{n}^{-1} \|} = \mathbf{0}_{\mathcal{A}}$. Finally, as $\dfrac{a_{n}^{-1}}{\| a_{n}^{-1} \|} \in \mathbb{S}(\mathcal{A})$ for each $n \in \mathbb{N}$, we conclude that $a$ is a left topological zero-divisor.

• Dear i.a.m, if you need further assistance with Problem (c), don’t hesitate to inform me. :) – Haskell Curry Jan 31 '13 at 9:33
• You might need to use the following fact: $\forall m,n \in \mathbb{N}: ~ a_{m}^{-1} - a_{n}^{-1} = a_{m}^{-1} a_{n}^{-1} (a_{n} - a_{m})$. – Haskell Curry Jan 31 '13 at 9:46
• I managed to get part c, can we conclude from part c that a must be a left topological zerodivisor? – i.a.m Feb 12 '13 at 5:09
• I managed to get part c, can we conclude from part c that a must be a left topological zerodivisor? – i.a.m Feb 12 '13 at 5:11
• @i.a.m: Yes. We can conclude from Part (c) that any $a \in \partial(\mathcal{G}(\mathcal{A}))$ must be a left topological zero-divisor. Please see the latest edit. – Haskell Curry Mar 5 '13 at 8:32

Hint for part (a): $b_n = a^{-1} (a b_n)$.