Exponential of a matrix, continuous function

Question

Take the Banach algebra $M_n(\mathbb{C})$ and consider the function that for a matrix $A \in M_n(\mathbb{C})$ associates the element $e^A = \sum\limits_{k = 0}^{\infty}\frac{A^k}{k!}.$ Is this function continuous?

My attempt: I know that the function "partial sum" $S_n: M_n(\mathbb{C}) \rightarrow M_n(\mathbb{C})$ given by $S_n(A) = \sum\limits_{k = 0}^{n}\frac{A^k}{k!}$ is continuous, because is polynomial. If a take matrices in a ball, for example $\{ A \in M_n(\mathbb{C}): ||A|| < K\}$ for some $K > 0 $ constant, then the partial sum converges uniformly to the exponential and by the uniform limit theorem i can conclude that the limit $A \mapsto e^A$ is continuous in that ball.

Answer 1

Let $p(A,B,n,m) = \displaystyle\sum_{u,v, \sum_i u_i = n,\sum_i v_i=m} \prod_{i=1} A^{u_i}B^{v_i}$ be the sum of the different permutations of $A^nB^m$ such that $$(A+B)^k = \sum_{m=0}^k p(A,B,k-m,m).$$

Now by the multiplicative property of $\|.\|$ : $$\|p(A,B,k-m,m)\| \le \sum_{u,v, \sum_i u_i = n,\sum_i v_i=m} \|\prod_{i=1} A^{u_i}B^{v_i} \|$$ $$\le \sum_{u,v, \sum_i u_i = n,\sum_i v_i=m} \|A\|^{k-m}\|B\|^m= {k \choose m} \|B\|^m\|A\|^{k-m}$$ so that $$\|(A+B)^k-A^k\| \le \sum_{m=1}^k\|p(A,B,k-m,m)\| \le \sum_{m=1}^k{k \choose m} \|B\|^m\|A\|^{k-m} = (\|A\|+\|B\|)^k-\|A\|^k.$$

For an analytic function $f(z) = \sum_{k=0}^\infty c_k z^k$ converging for $|z| < r$, with $\|A\|+\|B\| < r $: $$\|f(A+B)-f(A)\| \le \sum_{k=0}^\infty |c_k| \|(A+B)^k-A^k\| \le \sum_{k=0}^\infty |c_k|((\|A\|+\|B\|)^k-\|A\|^k) $$ $$= g(\|A\|+\|B\|)-g(\|A\|)$$ where $g(z) = \sum_{k=0}^\infty |c_k|z^k$ has the same radius of convergence $\ge r$,

hence it is continuous as $\|B\| \to 0$.

Answer 2

(This is an version of reuns answer with added clarity).

Let $\mathcal{A}$ be an associative algebra. Given $w=(A_1,\dots,A_n)\in\mathcal{A}^n$, we define its product $\operatorname{prod}(A_1,\dots,A_n)=A_1\dots A_n$ and given $B\in\mathcal{A}$, we define the multiplicity of $B$ in $w$ as $\operatorname{m}_B (A_1,\dots,A_n)=\#\{j\in\{1,\dots,n\}\mid A_j=B\}$, that is, the number of $B$'s in $w$.

Given $A,B\in\mathcal{A}$, let

$$ p(A,B,n,m) = \sum_{\substack{w\in\{A,B\}^{n+m}\\ \operatorname{m}_A w = n \\ \operatorname{m}_B w = m}} \operatorname{prod}w $$

be the sum of the different permutations of $A^nB^m$, so that

$$(A+B)^k = \sum_{m=0}^k p(A,B,k-m,m).$$

Suppose in addition that $\mathcal{A}$ is a normed algebra.

We have, by the multiplicative property of $\Vert\cdot\Vert$,

$$ \begin{align*} \|p(A,B,k-m,m)\| &=\left\lVert \sum_{\substack{w\in\{A,B\}^{k}\\ \operatorname{m}_A w = k-m \\ \operatorname{m}_B w = m}} \operatorname{prod}w \right\rVert\\ &\leq \sum_{\substack{w\in\{A,B\}^{k}\\ \operatorname{m}_A w = k-m \\ \operatorname{m}_B w = m}} \|\operatorname{prod}w\|\\ &\leq \sum_{\substack{w\in\{A,B\}^{k}\\ \operatorname{m}_A w = k-m \\ \operatorname{m}_B w = m}} \|A\|^{k-m}\|B\|^m\\ &= {k\choose m}\|A\|^{k-m}\|B\|^m; \end{align*} $$

so

$$\|(A+B)^k-A^k\| \le \sum_{m=1}^k\|p(A,B,k-m,m)\| \\\le \sum_{m=1}^k{k \choose m} \|B\|^m\|A\|^{k-m} = (\|A\|+\|B\|)^k-\|A\|^k.$$

Now, given an analytic function $f(z) = \sum_{k=0}^\infty c_k z^k$ converging for $|z| < r$ and if $\mathcal{A}$ is a Banach algebra (that is, a normed algebra which is also a Banach space), then the series $f(A)=\sum_{k=0}^\infty c_k A^k$ converges with $A\in\mathcal{A}$, $\|A\|< r$, since it converges absolutely in this case.

Finally, to see the continuity of $f(A)$, if $A,B\in\mathcal{A}$ are such that $\|A\|+\|B\| < r $, then

$$\|f(A+B)-f(A)\| \le \sum_{k=0}^\infty |c_k| \|(A+B)^k-A^k\| \le \sum_{k=0}^\infty |c_k|((\|A\|+\|B\|)^k-\|A\|^k) = g(\|A\|+\|B\|)-g(\|A\|)$$ where $g(z) = \sum_{k=0}^\infty |c_k|z^k$ converges for $|z|<r$.

Hence $f(A+B)\to f(A)$ as $\|B\| \to 0$, that is, we have continuity.