Homomorphisms on unitary complex Banach algebras are continuous Let $A$ be a complex Banach algebra and $h:A\to\mathbb{C}$ a homomorphism.
Let $h=0$ then the problem is trivial, so suppose that $h\ne 0$. Then there exists $x\in A$ with $h(x)\ne 0$ such that $h(x)=h(ex)=h(e)h(x)\implies h(e)=1$.
Now, we set $y=x-h(x)e$. Why does this belong to $\mathscr{N}(h)$? In a proof I have read it is claimed that $1=h(y y^{-1})=h(y)h(y^{-1})$, which is supposedly "impossible"; but I don't see how? And then it's claimed that this implies that $h(x)\in\sigma_{A}(x)$ and the conclusion is that $|h(x)|\le\|x\|$.
Would someone be able to explain why these final steps hold?
 A: First you need the step that $\sup|\sigma_A(x)|≤\|x\|$, which follows from $\sum_{n=0}^\infty (\frac{x}\lambda)^n$ converging in this case and then:
$$(\lambda\mathbb1-x)\frac1\lambda\sum_{n=0}^\infty\left(\frac{x}\lambda\right)^n=\sum_{n=0}^\infty \left(\frac x\lambda\right)^n-\sum_{n=1}^\infty \left(\frac{x}\lambda\right)^n=\mathbb1$$
and thus $(\lambda-x)$ is invertible whenever $\|x\|<\lambda$.
If you want to show that $h$ is continuous, you need to find an $L\in\mathbb R$ so that $|h(x)|≤L\|x\|$ for all $x$. We will show that you can actually choose $L$ to be $1$. Suppose that $|h(x)|>\|x\|$ for some $x$. The above statement gives you that $y=h(x)\mathbb1 - x$ is invertible, so you get that $1=h(\mathbb1)=h(yy^{-1})=h(y)h(y^{-1})$ and $h(y)$ cannot be $0$.
But: $h(y)=h(h(x)\mathbb1-x)=h(x)h(\mathbb1)-h(x)=h(x)-h(x)=0$. This is a contradiction, the map is not only continuous but a contraction.
A: To explain the argument that you read: You have $h(y)=0$. Suppose $y$ is invertible, $yy^{-1}=e$, this implies that $h(yy^{-1})=h(y)h(y^{-1})=h(e)=1$. This is impossible, so $y-h(x)e$ is not invertible. This implies that $h(x)$ is in the spectrum of $x$. The spectrum of $x$ is contained in the disc of radius $\|x\|$. This implies that $|h(x)|\leq \|x\|$.
https://en.wikipedia.org/wiki/Banach_algebra#Spectral_theory
