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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.

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  • $\begingroup$ The constant one function on the whole spectrum gives your operator back, and if it was idempotent then yes, you get one. Also, the spectrum of an idempotent is not too complicated... $\endgroup$ – András Bátkai Apr 14 '15 at 16:24
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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 ${1 \over 1- a/z}$.

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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 $$f(a) = \lim_{n \to \infty} g_n(a) = \lim_{n \to \infty} \left(g_n(0) + (g_n(1) - g_n(0)) a \right) = a$$

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  • $\begingroup$ I liked this answer too but the other one was more suitable for adaptation to what I wanted to do. $\endgroup$ – cyc Apr 15 '15 at 18:40

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