Holomorphic functional calculus and idempotents

Question

One of the applications of the holomorphic functional calculus is with regard to idempotents. For instance, if an element $a$ in a unital Banach algebra $A$ has spectrum contained in two balls, each of radius $\frac{1}{4}$ and centered at 0 and at 1 respectively, then by taking a holomorphic function that takes value 0 on the ball around 0 and takes value 1 on the ball around 1, one can apply the holomorphic functional calculus to produce an idempotent in $A$.

My question is: What happens if the original element $a$ was already an idempotent? Do we get $a$ itself after going through the same process?

I've looked at a few books that mention this application of the functional calculus but none of them mention this special case and I haven't been able to prove it myself. I hope it doesn't turn out to be a silly question.

Answer 1

Yes, you get your idempotent back.

You can prove it from the very definition of the functional calculus. Since $a^2=a$, we get $ker(1-a) \oplus ker(a) = A$ from elementary algebra. Now you can compute the inverse of $1-a+r e^{i \theta}$ explicitely in restriction to these two invariant subspaces.

$(1-a+re^{i\theta})^{-1}u= (1+re^{i\theta})^{-1}u \ $ if $u$ is in $ker(a)$

$(1-a+re^{i\theta})^{-1}u= (re^{i\theta})^{-1}u \ $ if $u$ is in $ker(1-a)$

Then plug this in the formula (with $r=1/4$)

$$f(a)u = {1\over 2\pi i} \int_{0}^{2\pi} (1-a+re^{i\theta})^{-1}u \ \ ire^{i\theta} d\theta$$

compute and you are done. More generally, if $a$ is the root of a polynomial with simple eigenvalues, then you just apply $f$ to the eigenvalues in some invariant subspace decomposition.

EDIT: oops I realize that you want an answer for all Banach algebras and not only for an algebra of operators. Then plug the following formula instead (with $z=1+re^{i\theta}$) $$ {1\over z-a} = {1 \over z}(1-a) + {1\over z-1} a$$ This formula can be obtained by a direct computation (using $a^2=a$). You can guess it by using a power series for ${1 \over 1- a/z}$.

Answer 2

Let $f$ be your function that is $0$ in a disk $D_0$ around $0$ and $1$ in a disk $D_1$ around $1$. There is (by Runge) a sequence of polynomials $g_n$ such that $g_n \to f$ uniformly on $D_0 \cup D_1$. Now for any polynomial $g(z) = c_0 + c_1 z + \ldots + c_n z^n$, $$g(a) = c_0 + (c_1 + \ldots + c_n) a = g(0) + (g(1) - g(0)) a$$ In particular $$f(a) = \lim_{n \to \infty} g_n(a) = \lim_{n \to \infty} \left(g_n(0) + (g_n(1) - g_n(0)) a \right) = a$$