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

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.