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 $\mathcal{A}$ be a unital Banach algebra with the norm $\|\cdot\|$ 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.

Dominated Convergence Theorem for Series

In this post, we will cover a classical but nevertheless powerful result about interchanging the order of limit and sum.

Let $G$ be an abelian group. We will adopt additive notation for the group operation. Assume that $G$ is endowed with a norm $\|\cdot\|$ with respect to which $G$ is complete.

Theorem. Let $(a^{(n)}_k)_{n,k=1}^{\infty} \subseteq G$ and $(M_k)_{k=1}^{\infty} \subseteq [0, \infty)$ satisfy the following conditions:

1. $\| a^{(n)}_k \| \leq M_k$ holds for each $n$ and $k$;
2. $(a^{(n)}_k)_{n=1}^{\infty}$ converges in $G$ as $k \to \infty$ for each $n$;
3. $\sum_{k=1}^{\infty} M_k$ converges.

Then $\sum_{k=1}^{\infty} a^{(n)}_k$ converges for each $n$, and $$\lim_{n\to\infty} \sum_{k=1}^{\infty} a^{(n)}_k = \sum_{k=1}^{\infty} \lim_{n\to\infty} a^{(n)}_k.$$

Remark. The completeness of $G$ is not necessary provided all the sums appearing in the equality converge in $G$.

Proof. For each $n$, let $S^{(n)} := \sum_{k=1}^{\infty} a^{(n)}_k$. Then for $p \lt q$, we have $$\left\| \sum_{k=p+1}^{q} a^{(n)}_k \right\| \leq \sum_{k=p+q}^{q} \| a^{(n)}_k\| \leq \sum_{k=p+1}^{q} M_k.$$ Since this bound converges to $0$ as $p, q \to \infty$, the partial sums of $S^{(n)}$ define a Cauchy sequence in $G$ for each $n$. Then the completeness of $G$ ensures that the series $S^{(n)}$ converges in $G$ for each $n$.

Define the sequence $(a_k)_{k=1}^{\infty}$ in $G$ by $a_k := \lim_{n\to\infty} a^{(n)}_k$ for each $k$. Then it is clear that $\| a_k \| \leq M_k$, and so, the sum $S := \sum_{k=1}^{\infty} a_k$ converges by the same reasoning. Finally, for each fixed $p$, $$\begin{align*} \| S^{(n)} - S \| &\leq \left\| S^{(n)} - \sum_{k=1}^{p} a_k \right\| + \left\| \sum_{k=1}^{p} a_k - S \right\| \\ &\leq \sum_{k=1}^{p} \| a^{(n)}_k - a_k \| + \sum_{k=p+1}^{\infty} \| a^{(n)}_k \| + \sum_{k=p+1}^{\infty} \| a_k \| \\ &\leq \sum_{k=1}^{p} \| a^{(n)}_k - a_k \| + 2 \sum_{k=p+1}^{\infty} M_k. \end{align*}$$ Letting $\limsup$ as $n \to \infty$, we obtain the bound $$\limsup_{n\to\infty} \| S^{(n)} - S \| \leq 2 \sum_{k=p+1}^{\infty} M_k.$$ Since the left-hand side does not depend on $p$, letting $p \to \infty$ shows that the limsup is in fact $0$, proving $S^{(n)} \to S$ in $G$ as $n \to \infty$. This conclusion is equivalent to the desired assertion, hence the proof is done. $\square$

Example. We show that $$\sum_{k=1}^{\infty} \frac{2^k}{k} = 0$$ in $\mathbb{Q}_2$. Indeed, for $n \geq 1$ we have $$\begin{align*} 0 = \frac{1 - (1 - 2)^{2^n}}{2^n} &= \frac{1}{2^n} \sum_{k=1}^{\infty} (-1)^{k-1} \binom{2^n}{k} 2^k \\ &= \sum_{k=1}^{\infty} (-1)^{k-1} \binom{2^n - 1}{k - 1} \frac{2^k}{k} \\ &= \sum_{k=1}^{\infty} \left[ \prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \right] \frac{2^k}{k}. \end{align*}$$ It is easy to check that the $|\cdot|_2$-norm of the $k$th term is uniformly bounded by $2^{-k}$ for all $n$ and $k$, and $\prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \to 1$ in $\mathbb{Q}_2$ as $n \to \infty$ for each $k$. So by DCT, it follows that $$0 = \sum_{k=1}^{\infty} \left[ \lim_{n \to \infty} \prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \right] \frac{2^k}{k} = \sum_{k=1}^{\infty} \frac{2^k}{k}.$$

Remark. The above result can be interpreted in terms of logarithm as follows: Note that the logarithm $\log(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} z^n$ converges in $\mathbb{Q}_2$ whenever $|z|_2 < 1$, and moreover, $$\log(1+z) + \log(1+w) = \log((1+z)(1+w)),$$ which initially holds as an identity between formal power series, extends to all $z, w \in \mathbb{Q}_2$ with $|z|_2 < 1$ and $|w|_2 < 1$. So, $$\sum_{n=1}^{\infty} \frac{2^n}{n} = - \log(1-2) = - \log (-1) = - \frac{1}{2} \log((-1)^2) = -\frac{1}{2}\log 1 = 0.$$