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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.

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  • $\begingroup$ anything unclear ? $\endgroup$ – reuns Oct 17 '16 at 12:12
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  • Let $p(A,B,n,m) = \displaystyle\sum_{u,v, \sum_i u_i = n,\sum_i v_i=m} \prod_{i=1} A^{u_i}B^{v_i}$ 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 $\|B\| \to 0$.

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