Convergence of $\prod_{k=1}^n \left( I_d + \frac{1}{n} A\left(\frac{k}{n}\right) \right)$ for $A : [0, 1] \rightarrow \mathcal{M}_d(\mathbf{R})$

Question

The problem

Let $d \geq 1$ and $A : [0, 1] \rightarrow \mathcal{M}_d(\mathbf{R})$ be a continuous function. For $n \geq 1$, define:

$$ E_n := \prod_{k=1}^n \left( I_d + \frac{1}{n} A \left( \frac{k}{n} \right) \right) $$

The goal is to study the convergence of $(E_n)_{n \geq 1}$. I would like to find when (under which conditions on $A$) this sequence converges, to which limit, and when it does not converge.

My try

My first idea is to show that:

$$ E_n \underset{n \to +\infty}{\longrightarrow} \exp \left( \int_0^1 A \right) $$

I have succeeded to prove it for $d = 1$. It consists in taking the logarithm of $E_n$, use the inequalities $x - \frac{x^2}{2} \leq \ln(1+x) \leq x$ for $x \in [0, 1]$ and finally the squeeze theorem and Riemann sums.

The result is also true when $A$ is a constant matrix.

But I have a hard time to generalize the proof with logarithms. I know one can define the logarithm of matrices not too far from $I_d$, but nothing about the additivity of this logarithm.

Answer 1

For the convergence, we write

$$ X_{n,i} := \frac{1}{n^i} \sum_{1 \leq k_1 < \cdots < k_i \leq n} A\bigl(\tfrac{k_1}{n}\bigr)\cdots A\bigl(\tfrac{k_i}{n}\bigr). $$

(Here, we set $X_{n,0} := I_d$ and $X_{n,i} := 0$ for $i > n$.) Then we may write the product as

$$ E_n := \biggl(I_d + \frac{A\bigl(\tfrac{1}{n}\bigr)}{n} \biggr) \cdots \biggl(I_d + \frac{A\bigl(\tfrac{n}{n}\bigr)}{n}\biggr) = \sum_{i=0}^{\infty} X_{n,i}. $$

Since the operator norm of $X_{n,i}$ admits the uniform bound

$$ \| X_{n,i} \| \leq \frac{1}{i!} \|A\|_{\sup}^i $$

with $\|A\|_{\sup} := \sup_{0 \leq t \leq 1} \|A(t)\|$, Weierstrass M-test shows that $E_n$ converges provided $X_{n,i}$ converges as $n\to\infty$ for each $i$. But if we write $\mathcal{T}_i := \{ (t_1, \cdots, t_i) : 0 \leq t_1 \leq \cdots \leq t_i \leq 1\}$, then

$$ \lim_{n\to\infty} X_{n,i} = \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i, $$

the desired conclusion follows. Moreover, we obtain:

$$ \lim_{n\to\infty} \biggl(I_d + \frac{A\bigl(\tfrac{1}{n}\bigr)}{n} \biggr) \cdots \biggl(I_d + \frac{A\bigl(\tfrac{n}{n}\bigr)}{n}\biggr) = \sum_{i=0}^{\infty} \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i. $$

If $\{ A(t) \}_{t \in [0, 1]}$ commutes, then the above integral reduces to

$$ \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i = \frac{1}{n!} \biggl( \int_{0}^{1} A(t) \, \mathrm{d}t \biggr)^i $$

and thus the limit of $E_n$ becomes the matrix exponential $\exp\bigl( \int_{0}^{1} A(t) \, \mathrm{d}t \bigr)$. But we do not expect this to happen in general.

Answer 2

The statement is wrong. Reason for $d=2$ and consider the matrices $$ T := \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \quad S : = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$ Notice that we do not have $\exp(T + S) = \exp(T)\exp(S)$ by a simple computation, which can be done noticing that $$ S^2 = -I_2 \text{ and } T^n = \begin{pmatrix} 1 & n\\0 &1 \\ \end{pmatrix}$$ for all integers $n$. Now if we relax the condition on $A$ to be only piecewise continuous, it is then clear that the statement fails: take $A(t) = 2T$ on $[0,1/2]$ and $A(t) = 2S$ on $(1/2,1]$ and use the fact that $$ \lim_{n \to \infty} \left(I_2 + \frac{M}{n} \right)^n = \exp(M) $$ for all matrices $M$. Now the idea is just to approximate this piecewise constant function by an affine function. Reason by contradiction and suppose $E_n$ converges for all continuous functions to $\exp(\int_0^1 A)$. Let $N$ be an integer indexing the sequence of continuous function $(A_N)$ where $A_N$ is defined to be constant equal to $2T$ on $[0,1/2]$, an affine function connecting $2T$ to $2S$ on $[1/2, 1/2 + 1/N]$ and equal to $S$ on $[1/2 + 1/N, 1]$. Then the product $E^N_{2n}$ associated with this function $A_N$ can be written as $$E^N_{2n} = \left( I_2 + \frac{T}{n} \right)^n \cdot \; \prod_{k=n}^{n+ \lfloor 2n/N \rfloor } (I_2 + (1/2n)A_N(k/n)) \cdot (I_2 + S/n)^{n - \lfloor 2n/N \rfloor}$$ Both left and right side of this product converge to $\exp(T)$ and $\exp(S(1-2/N))$ respectively by easy estimations. Therefore since $E_n^N$ converges, it must be that the term in the middle converges as well, say towards a matrix $M_N$. We hence have the relation $$\exp\left(\int_0^1 A_N\right) = \exp(T)M_N\exp(S(1-2/N)) $$ The norm (say the infinity norm on the coefficients) of $M_N - I_2$ is the limit of the norm of the product in the middle minus the identity, this norm is bounded by $(1 + 1/2n)^{\lfloor2n/N\rfloor} -1$ (expand the product without using commutativity and use the triangle inequality), hence $\|M_N - I_2 \| \leq e^{1/N} -1 $, thus $M_N$ converges towards $I_2$ as $N$ approaches infinity. Now you just have to let $N \to \infty$ in the formula above to derive a contradiction.