I have to prove the following:
Let $A$ be a Banach algebra with unit $e.$ If there exists $M>0$ such that $\Vert xy \Vert \leq M \Vert yx \Vert$ for all $x, y \in A,$ then the function $\lambda \in \mathbb{C} \mapsto \exp(-\lambda x)y \exp(\lambda x) \in A$ is constant in $\mathbb{C}$ and, as a consequence, $A$ is commutative.
This is what I've done:
Preliminary work
Recall that, for each $x \in A,$ we define $$ \exp(x) := \frac{1}{2\pi i} \int_\Gamma e^z (ze-x)^{-1} \, dz := \frac{1}{2\pi i} \int_a^b e^{\Gamma(t)} \Gamma'(t) (\Gamma(t)e-x)^{-1} \, dt,$$ where $\Gamma \colon [a, b] \to \mathbb{C}$ is any path that goes around $\sigma(x)$ in $\mathbb{C}$ and the second integral is a Bochner integral.
Notice that $$(ze-x)^{-1} = \frac{1}{z} \left( e-\frac{x}{z}\right)^{-1} = \sum_{n=0}^\infty \frac{x^n}{z^{n+1}}$$ provided that $\Vert x/z \Vert < 1$ or, equivalently, $|z|>\Vert x \Vert.$ And since this series converges uniformly on $\text{supp } \Gamma$ (seeing $z$ as the variable), we have $$ \exp(x) = \frac{1}{2\pi i} \int_\Gamma e^z \sum_{n=0}^\infty \frac{x^n}{z^{n+1}} \, dz = \sum_{n=0}^\infty x^n \left( \frac{1}{2\pi i}\int_\Gamma \frac{e^z}{z^{n+1}} \, dz \right) = \sum_{n=0}^\infty \frac{x^n}{n!};$$ in the last equal sign we have used Cauchy's formula for derivatives and the fact that all the derivatives of the complex exponential at 0 equal 1.
This implies that $\exp(x+y)=\exp(x)\exp(y)$ for any $x, y \in A$ with $xy=yx.$ Indeed, in that case, we can use Newton's binomial theorem to obtain that $$\exp(x+y) = \sum_{n=0}^\infty \frac{(x+y)^n}{n!} = \sum_{n=0}^\infty \sum_{k=0}^n {n \choose k}x^k y^{n-k} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{x^k}{k!} \frac{y^{n-k}}{(n-k)!},$$ and because $\sum_{n=0}^\infty x^n/n!$ and $\sum_{m=0}^\infty y^m/m!$ converge absolutely, this equals $$\left( \sum_{n=0}^\infty \frac{x^n}{n!} \right) \left( \sum_{m=0}^\infty \frac{y^m}{m!} \right) = \exp(x)\exp(y).$$
Proof
Fix $x,y \in A$ and call $F \colon \mathbb{C} \to A$ the function given by $F(\lambda) = \exp(-\lambda x) y \exp(\lambda x).$ According to our preliminary work, we have that $$F(\lambda)= \left(\sum_{n=0}^\infty \frac{(-1)^n \lambda^n x^n y}{n!} \right) \left( \sum_{m=0}^\infty \frac{\lambda^m x^m}{m!}\right),$$ and since these series converge absolutely, $$F(\lambda)=\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k\lambda^nx^kyx^{n-k}}{k!(n-k)!}$$ for all $\lambda \in \mathbb{C}.$
Now, if $x^* \in A^*,$ the function $x^* \circ F \colon \mathbb{C} \to\mathbb{C}$ verifies $$ x^* \circ F(\lambda) = \sum_{n=0}^\infty \left[ \sum_{k=0}^n \frac{(-1)^kx^*(x^kyx^{n-k})}{k!(n-k)!} \right] \lambda^n$$ for all $\lambda \in \mathbb{C};$ we have proven that $x^* \circ F$ is given by a power series and hence it is an entire function.
Observe that $x^* \circ F$ is bounded: for any $\lambda \in \mathbb{C},$ it holds that $|x^* \circ F(\lambda)| \leq \Vert x^* \Vert \Vert \exp(-\lambda x) y \exp(\lambda x) \Vert \leq \Vert x^* \Vert M \Vert y \exp(\lambda x) \exp(-\lambda x) \Vert = \Vert x^* \Vert M$ (because $-\lambda x$ and $\lambda x$ commute). By Liouville's theorem, we conclude that $x^* \circ F(\lambda) =x^*(y)$ for all $\lambda \in \mathbb{C}.$ Therefore the coefficient of $\lambda$ in the power series expansion of $x^* \circ F$ must be 0; that is, $x^*(xy) = x^*(yx).$ This is true for any $x^* \in A^*,$ and $A^*$ separates points in $A$ (as a consequence of the Hahn-Banach theorem), so $xy = yx.$
I'd like to know if my argument is correct and if this fact can be proven without using the power series expansion.