I've come across the following question and can't find the mistake in my counterexample.
Let $A$ be a square matrix whose elements are real valued continuous functions of time on the interval $t_0 \leq t \leq t_1$ and let the sequence of matrices $M_k$ be defined recursively by $$ M_0 = I$$ $$ M_k = I + \int_{t_0}^t A(s) M_{k-1}(s) ds;\ k \geq 1$$ Show that $(M_k)_{k \in \mathbb{N}}$ (componentwise) converges uniformly on the given interval.
My counter example:
Let $t_0 = 0, t = t_1 = 1, A(s) := I$.
Then
$$M_k = (k+1)I$$
Proof by Induction on k:
- $k=0$: $M_0 = I$ is by definition.
- $(k-1) \rightarrow k:$ $$ M_k = I + \int_{t_0}^{t_1} A(s) M_{k-1}(s) ds$$ $$ = I + \int_{t_0}^{t_1} I \cdot M_{k-1}(s) ds$$ $$ = I + \int_{t_0}^{t_1} M_{k-1}(s) ds$$ By induction hypothesis $$ = I + \int_{t_0}^{t_1} k I ds$$ $$ = I + k \int_{t_0}^{t_1} I ds$$ $$ = I + k I (1-0) = (k+1) I$$
Therefore, $M_k$ diverges.
Where is my mistake in this proof?
The recursive definition of $M_k$ actually means $M_k(t)=I+\int_{t_0}^t A(s)M_{k-1}(s)ds$ for all $t\in[t_0,t_1]$ and $k\ge 1$, and so you are not allowed to substitute $t=t_1$ and proceed as such. Otherwise, you are simply putting $M_k(t_1)$ instead of $M_k(s)$ in the recursive definition.
Indeed, as @Sangchul Lee pointed out in a comment, in your case $\lim_{k\to\infty}M_k(t)=e^{(t-t_0)}I$, where the limit is of course component-wise. This limit is famously uniform (on compact sets).