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This question is from Differential calculus by Henri Cartan:

Suppose that $fg = gf$ for $f,g\in L(E,E)$ where $E$ is a banach space and $\exp(f) = \displaystyle{\sum_{n\geq0} \frac{f^n}{n!}}$. Show that $\exp(f+g)=\exp(f)\exp(g)$.

I am learning this on my own, which is why I'm so confused. Would someone please give me a hint?

PaulichenT
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Mina
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1 Answers1

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As PaulichenT mentioned this link can help you, but I would like to give a different answer. We have that \begin{align*} \exp(f) \exp(g) &= \left(\sum_{n=0}^{\infty} \frac{f^n}{n!}\right)\left(\sum_{n=0}^{\infty} \frac{g^n}{n!}\right)=\sum_{n=0}^{\infty} c_n \end{align*} When \begin{align*} c_n&=\sum_{k=0}^{n}\frac{f^{n-k}}{(n-k)!}\frac{g^k}{k!}\\[2mm] &=\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{(n-k)!k!}f^{n-k}g^k\\[2mm] &=\frac{1}{n!}(f+g)^n \end{align*} thus \begin{align*} \exp(f) \exp(g) &=\sum_{n=0}^{\infty} c_n=\sum_{n=0}^{\infty}\frac{(f+g)^n}{n!}=\exp(f+g) \end{align*}

Zaragosa
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