Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$.
If $u_A:=\sum_{n=0}^\infty \frac{i^nA^nu}{n!}$ is converge, $u_A=e^{iA}u$.
I cannot prove this. I will appreciate your help with this situation.
2025-01-13 05:43:42.1736747022
Exponential of self-adjoint operator
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$\sup_{\|u\| \leq 1}\| \sum\limits_{k=n}^{m} \frac {i^{n}A^{n}u} {n!}\| \leq \sum\limits_{k=n}^{m} \frac {\|A\|^{n}} {n!}\to 0$ as $m >n \to \infty$. Hence the sereies converges in operator norm (and sum is defined as $e^{iA}$).