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} \} $.
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 $.
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 $
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 $
(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}_{+} $.
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.