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I want to prove that the function $ \exp\colon M_n(\mathbb{C})\to \mathrm{GL}_n(\mathbb{C}) $ is continuous under standard matrix norm $$ \lVert A\rVert=\sup_{\lVert x\rVert=1}\lVert Ax\rVert. $$ Wikipedia says that it follows from the inequality $$ \lVert e^{X+Y}-e^X\rVert\leqslant \lVert Y\rVert e^{\lVert X\rVert}e^{\lVert Y\rVert}, $$ and I understand why, but I don't quite follow how to get this inequality. Could somebody explain that?

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Let $$ p:M_n(\Bbb C )^k\to M_n(\Bbb C ),\quad (X_1,X_2,\ldots ,X_n)\mapsto X_1\cdot X_2\cdots X_n\tag1 $$ the ordered product of a vector of matrices, and $$ c_X:\{X,Y\}^k\to \{0,\ldots ,k\}\tag2 $$ is a function that count the number of coordinates of a vector in $\{X,Y\}^k$ that are equal to $X$. Then we have that $$ (X+Y)^k=\sum_{v\in \{X,Y\}^k}p(v)=\sum_{j=0}^k\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)\tag3 $$ And if $X$ and $Y$ commute then $$ \sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)=\binom{k}{j}X^jY^{k-j}\tag4 $$ Then from $\mathrm{(3)} $ we have that $$ \begin{align*} \|(X+Y)^k-X^k\|&\leqslant \left\|\sum_{j=0}^k\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)-X^k\right\|\\ &=\left\|\sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)\right\|\\ &\leqslant \sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}\|p(v)\|\\ &\leqslant \sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}\|X\|^j\|Y\|^{k-j}\\ &=\sum_{j=0}^{k-1}\binom{k}{j}\|X\|^j\|Y\|^{k-j}\\ &=\sum_{j=0}^{k-1}\binom{k}{j}\|X\|^j\|Y\|^{k-j}+\|X\|^k-\|X\|^k\\ &=(\|X\|+\|Y\|)^k-\|X\|^k\tag5 \end{align*} $$ where in the third inequality we used implicitly the inequality $\|AB\|\leqslant \|A\|\|B\|$ for any square matrices $A$ and $B$. Then you have that $$ \begin{align*} \|e^{X+Y}-e^X\|&=\left\|\sum_{k\geqslant 0}\frac{(X+Y)^k-X^k}{k!}\right\|\\ &\leqslant \sum_{k\geqslant 0}\frac{\|(X+Y)^k-X^k\|}{k!}\\ &\leqslant \sum_{k\geqslant 0}\frac{(\|X\|+\|Y\|)^k-\|X\|^k}{k!}\\ &=e^{\|X\|+\|Y\|}-e^{\|X\|}\\ &=e^{\|X\|}(e^{\|Y\|}-1)\tag6 \end{align*} $$ And $$ e^c-1\leqslant ce^c\iff \sum_{k\geqslant 0}\frac{c^{k+1}}{(k+1)!}\leqslant \sum_{k\geqslant 0}\frac{c^{k+1}}{k!}\tag7 $$ clearly holds for $c\geqslant 0$. Then $\mathrm{(6)} $ and $\mathrm{(7)} $ prove your inequality.

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    $\begingroup$ Why the very first inequality holds? $\endgroup$ – Kumquat Jan 18 at 0:57
  • $\begingroup$ @Kumquat I added more details $\endgroup$ – Masacroso Jan 18 at 5:06

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