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Saturday, October 15, 2022

Spitzer's Formula

The following identity is an abstract version of the Spitzer's formula introduced in Wendel's 1958 paper [1] and the proof thereof, with some simplifications.

Theorem. Let A be a unital Banach algebra with the norm and unit 1. Let \mathbf{P} be a bounded linear operator on \mathcal{A} satisfying the following conditions:

  1. \mathbf{P} is a projection, that is, \mathbf{P}^2 = \mathbf{P}.
  2. Both \operatorname{im}\mathbf{P} and \operatorname{im}(\mathbf{I}-\mathbf{P}) are closed subalgebras of \mathcal{A}.
  3. x and \mathbf{P}x commute for any x \in \mathcal{A}.
Then for any a \in \mathcal{A} and with the multiplication operator \mathbf{M}_a defined by \mathbf{M}_a(x) = ax, we have \sum_{n=0}^{\infty} t^n (\mathbf{P}\mathbf{M}_a)^n 1 = \exp\left(\sum_{n=1}^{\infty} \frac{t^n}{n} \mathbf{P}(a^n) \right) \tag{*} for any scalar t within the radius of convergence of both sides.

Proof. Write \mathbf{Q} = \mathbf{I} - \mathbf{P} for simplicity. We first note that, for any scalars a_n's and b_n's and x \in \mathcal{A}, \mathbf{Q} \left[ \sum_{n=0}^{\infty} a_n (\mathbf{P}x)^n \right] = \mathbf{Q}a_0 \qquad\text{and}\qquad \mathbf{P} \left[ \sum_{n=0}^{\infty} b_n (\mathbf{Q}x)^n \right] = \mathbf{P}b_0 \tag{1} provided the respective series converges. Now let t be sufficiently small so that all the series appearing in \text{(*)} converges absolutely, and let g denote the left-hand side of \text{(*)}: g = \sum_{n=0}^{\infty} t^n (\mathbf{P}\mathbf{M}_a)^n 1 = (\mathbf{I} - t \mathbf{P}\mathbf{M}_a)^{-1} 1 Hence, g is uniquely determined by the equation (\mathbf{I} - t \mathbf{P}\mathbf{M}_a) g = 1. \tag{2} In light of the above observation, it suffices to prove that the right-hand side of \text{(*)} satisfies the condition \text{(2)}. Indeed, let g^* denote the right-hand side of \text{(*)}: g^* = \exp\left(\sum_{n=1}^{\infty} \frac{t^n}{n} \mathbf{P}(a^n) \right) = \exp\left(- \mathbf{P} \log(1 - ta) \right) Then we get \begin{align*} (\mathbf{I} - t \mathbf{P}\mathbf{M}_a) g^* &= \mathbf{Q}g^* + \mathbf{P}(\mathbf{I} - t\mathbf{M}_a)g^* \\ &= \mathbf{Q}g^* + \mathbf{P}((1 - ta)g^*) \\ &= \mathbf{Q}\exp\left( -\mathbf{P} \log(1 - ta) \right) + \mathbf{P} \exp\left( \mathbf{Q} \log(1 - ta) \right) \\ &= \mathbf{Q}1 + \mathbf{P}1 \\ &= 1. \end{align*} In the third step, we utilized the fact that e^x e^y = e^{x+y} provided x and y commute. Then \text{(1)} is used in the fourth step. Therefore g^* = g. \square

References.

[1] Wendel, James G. (1958). "Spitzer's formula: A short proof". Proceedings of the American Mathematical Society. 9 (6): 905–908. doi:10.1090/S0002-9939-1958-0103531-2. MR 0103531.