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How to show that if A- normal operator in H, where H-separable Hilbert space, $B_n = (Id_H + \frac{A}{n}) ^ n$ converges in the norm of $ || *||_{L (H, H)} $ to $ expA $, using the spectral theorem?

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  • $\begingroup$ Do you know how to show it in the simple case $H=\mathbb{R}$? $\endgroup$ – Ian May 19 '15 at 21:55
  • $\begingroup$ No, unfortunately I do not know. You could show?Please. $\endgroup$ – user241965 May 20 '15 at 14:49
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Use a branch of $\log(1+z)$ with a branch cut along the negative real axis. For large enough $n$, $1+\lambda/n$ is within $\{ z\in\mathbb{C} : |z-1| \le 1/2\}$ and $$ \left(I+\frac{1}{n}A\right)^{n} = \int_{\sigma(A)}\left(1+\frac{\lambda}{n}\right)^{n}dE(\lambda) \\ = \int_{\sigma(A)}e^{n\log(1+\lambda/n)}dE(\lambda) \\ \rightarrow \int_{\sigma(A)}e^{\lambda}dE(\lambda). $$ The convergence of the integrand is uniform in $\lambda\in\sigma(A)$ because the spectrum is bounded. So above sequence of operators converges in $\mathcal{L}(H)$.

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