Can anyone help me by providing a detailed verification of the following theorem?
Let $\mathcal{A}$ be a Banach algebra. If there exists $M<+\infty$ so that $$\Vert a \Vert\Vert b \Vert\leq M \Vert ab \Vert,$$ $$(a,b\in\mathcal{A})$$ then $\mathcal{A}$ is isometrically isomorphic with $\mathbb{C}.$
Assume that $A$ is unital (otherwise we can take $A$ to be zero-dimensional). We need to prove that every non-zero element $x$ is invertible and then apply Gelfand-Mazur theorem. The set of non-invertible elements is closed. If it contains any non-zero element $x$ then its boundary also contains some non-zero element, call it $y$. Since $y$ is a boundary point, it is a limit of a sequence of invertible elements $(y_n)$. From our assumptions we gather that $\|y_n\| \cdot \|y_n^{-1}\| \leqslant M$, so the sequence $(\|y_n^{-1}\|)$ is bounded; this means that $y = \lim_{n \to \infty} y_n$ is invertible as well and this is a contradiction. The explanation is that $\|e - y_n^{-1}y\| \leqslant \|y_n^{-1}\| \cdot \|y_n -y\| < 1$ for sufficiently large $n$, so $y_n^{-1}y$ is invertible for such $n$.