If $\{T_n\}$ converges in operator norm to $T$ does it follow that $\exp(T_n)$ converges to $\exp(T)$ at least in finite dimensions? I can handle this when the operators commute but not in general.
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3Is $\exp$ continuous? – edm Oct 10 '17 at 07:03
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Even Lipschitz continuous from MVT – A.Γ. Oct 10 '17 at 07:23
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Do you define $\exp(T)$ by means of the series? How about polynomials? Try to show the norm convergence $p(T_n)\to p(T)$. – Yurii Savchuk Oct 10 '17 at 07:33
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Even in one dimension, the exponential map is not Lipschitz. – Kavi Rama Murthy Oct 10 '17 at 08:58
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$$ \exp(T_n)-\exp(T)=\exp((1-t)T_n)\exp(tT)|_{t=0}^{1} \\ = \int_{0}^{1}-\exp((1-t)T_n)T_n\exp(tT)+\exp((1-t)T_n)T\exp(tT)dt \\ = \int_{0}^{1}\exp((1-t)T_n)(T-T_n)\exp(tT)dt $$ Therefore, $$ \|\exp(T_n)-\exp(T)\| \le \int_{0}^{1}\exp((1-t)\|T_n\|)\|T-T_n\|\exp(t\|T\|)dt. $$
Disintegrating By Parts
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The estimates I had for $||A^n-B^n||$ don't hold for non-commuting operators but a simple application of Dominated Convergence Theorem proves continuity of the exponential map in the operator norm, even in infinite dimensions. Anyway, thanks for the comments.
Kavi Rama Murthy
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Let $S = T_n-T$ so that $\|S_n\| \to 0$ $$\exp(T+S_n)-\exp(T) = \sum_{k=1}^\infty \frac{1}{k!}((T+S_n)^k-T^k)$$ Use the non-commutative binomial sum $$(T+S_n)^k = \sum_{a \in \{0,1\}^k } \prod_{i=1}^k T^{a_i} S_n^{1-a_i}$$ For $\|S_n\| \le \|T\|$ $$ \|(T+S_n)^k -T^k\| \le \sum_{a \in \{0,1\}^k,a \ne \{1\}^k } \|\prod_{i=1}^k T^{a_i} S_n^{1-a_i}\|\quad \le \quad 2^k \|S_n\|\|T\|^{k-1}$$ $$\|\exp(T+S_n)-\exp(T)\| \le \|S_n\| \sum_{k=1}^\infty \frac{2^k \|T\|^{k-1}}{k!}= \|S_n\| \frac{e^{2\|T\|}-1}{\|T\|}$$
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