Let $T$ be a linear operator from $H$ to itself. If we define $\exp(T)=\sum_{n=0}^\infty \frac{T^n}{n!}$ then how do we prove the function $f(\lambda)=exp(\lambda T)$ for $\lambda\in\mathbb{C}$ is differentiable on a Hilbert space?
Exponential operator on a Hilbert space
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analysis
derivatives
hilbert-spaces
operator-theory
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0I'm not sure if I understand the question. Are you saying that $T$ is a linear operator from a Hilbert Space to itself? And you want to know whether the map $f$ in question is a differentiable map from the complex numbers to the space of linear operators on the Hilbert Space? – 2012-11-10
1 Answers
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$\frac{f(\lambda)-f(0)}{\lambda}=\frac{\exp(\lambda T)-Id}{\lambda} = \frac1\lambda\left( \sum_{n=1}^{\infty} \frac{\lambda^nT^n}{n!} \right) = \sum_{n=1}^{\infty} \frac{\lambda^{n-1}T^n}{n!}$
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0Apparently there was no need :-) – 2012-11-10