Power Series: Derivative

Question

Given a Banach space $E$.

Consider a series: $$|t|\leq R:\quad\sum_{k=0}^\infty A_k t^k\quad(A_k\in E)$$

Is there an elegant proof of: $$\left(\sum_{k=0}^\infty A_k t^k\right)'=\sum_{k=0}^\infty A_k kt^{k-1}$$

Already, the radii of convergence agree: $$\limsup_{k\to\infty}\sqrt[k]{(k+1)A_{k+1}}=\limsup_{k\to\infty}\sqrt[k]{kA_k}$$

Moreover, the series converge uniformly on inner disks as: $$\|\sum_{k=K+1}^\infty A_kt^k\|\leq\sum_{k=K+1}^\infty\|A_k\|\cdot|t|^k\leq\sum_{k=K+1}^\infty\|A_k\|\cdot r^k\to0\quad(r<R)$$

Answer 1

If $F : \Omega\rightarrow E$ is a vector function on an open domain in $\mathbb{C}$ into a Banach space $E$, then $F$ is holomorphic iff it is weakly holomorphic. In that case $(x^{\star}\circ F)'(\lambda)= x^{\star}\circ F'(\lambda)$ for all $x^{\star}\in E^{\star}$.

Your power series is weakly holomorphic within the radius of convergence, with $$ \begin{align} \left(x^{\star}\circ\sum_{k=0}^{\infty}\lambda^{k}A_{k}\right)' & =\left(\sum_{k=0}^{\infty}\lambda^{k}x^{\star}(A_{k})\right)' \\ & =\sum_{k=1}^{\infty}k\lambda^{k-1}x^{\star}(A_{k}) \\ & = x^{\star}\circ\sum_{k=1}^{\infty}k\lambda^{k-1}A_{k}. \end{align} $$

Answer 2

If you know the result in the scalar case you can use Hahn-Banach to extend to the Banach case.