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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?

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What is the variable now, $\lambda$? And the function, $\lambda\mapsto\exp(\lambda T)$ for a fixed $T$? –  Berci Nov 10 '12 at 20:54
    
i fixed the question –  john Nov 10 '12 at 20:58
    
I'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? –  Braindead Nov 10 '12 at 21:09

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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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Plus the fact that $f(\lambda + \mu) = f(\lambda)f(\mu)$. –  commenter Nov 10 '12 at 21:21
    
Yes, yes, I wanted to add it. –  Berci Nov 10 '12 at 21:33
    
Apparently there was no need :-) –  commenter Nov 10 '12 at 21:35

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