If $A \in B(V)$ where $V$ is a Banach space, and $B(V)$ is set of all bounded linear operators on $V$, then show if $lim_{n→\infty} A_n=A$, then $lim_{n→\infty}(I+\frac{A_n}{n})^n = exp(A)$
I am very new in learning matrix exponentials and norm linear spaces, and but I can't approach.
Obviously $(I+\frac{A_n}{n})^n$ is bounded .
A small help is appreciated .
Here's one way to prove it:
Step 1): For each fixed complex number $z$ have have $(1 + \frac{z}{n})^n \to \exp(z)$; this is simply Euler's formula.
Step 2): If $|z| \le R$ for a real number $R \ge 0$, then $$ \big|(1 + \frac{z}{n})^n\big| \le \big(1 + \frac{R}{n}\big)^n \to \exp(R). $$ This shows that the sequence $\big((1 + \frac{z}{n})^n\big)_{n \in \mathbb{N}}$ is uniformly bounded for $z$ in any fixed compact subset of the complex plane.
Hence, the convergence $(1 + \frac{z}{n})^n \to \exp(z)$ is uniform on compact subsets of $\mathbb{C}$ by Step 1) and Montel's theorem.
Step 3): General observation:
If $(f_n)$ is a sequence of entire functions which converges to $0$ uniformly on compact subsets of $\mathbb{C}$, then for each number $R \ge 0$, $$ \sup\big\{ \|f_n(C)\|: \, C \in B(V), \; \|C\| \le R \big\} \to 0 \quad \text{as } n \to \infty. $$ Indeed, this follows from the Cauchy formula representation of the functional calculus, i.e., $$ f_n(C) = \frac{1}{2\pi i} \oint_{|z| = {R+1}} f_n(z) (z-C)^{-1} \, dz. $$ (Use that boundedness of the resolvent $(z-C)^{-1}$ on the circle $|z| = R+1$ follows from $\|C\| \le R$ and from the Neumann series representation of the resolvent.
Step 4): Use a $2\varepsilon$-argument: For each index $n$, we have $$ \left\|(1 + \frac{A_n}{n})^n - \exp(A)\right\| \le \left\|(1 + \frac{A_n}{n})^n - \exp(A_n)\right\| + \left\| \exp(A_n) - \exp(A) \right\|. $$ The first summand converges to $0$ according to Steps 2) and 3) above, and the second summand converges to $0$ due to the continuity of the operator exponential function.