I am trying to take a derivative w.r.t $z\in\mathbb{C}$ of the following map:
$z\mapsto \sum_{j=0}^{\infty}\lambda_{j} (T(\psi+zh))_{j}$
where $(\lambda_{j})$ is a bounded sequence, $T$ is a nonlinear operator on a complex Banach space (with $l^{1}$ norm, functions analytic on some open disc radius $r$), and both $\psi$ and $h$ belong to that space.
How do I take a derivative here? Does the usual chain rule apply? What confuses me is that I have to differentiate w.r.t. a variable, and then it's multiplied by a function… So by the chain rule I probably cannot say that the inner derivative is just $h$?
What makes me very confused is that I'm looking for a "normal" complex derivative, but not just of a function of a complex variable, but of an operator. Thanks for any help!
Yes, the chain rule applies here, assuming $T$ is at least Gateaux differentiable. The derivative you consider is the direct analog of directional derivatives of multivariable calculus, $D_v f = \frac{d}{dt}f(x+tv)$ where $x,v$ are elements of vector space and $t$ is a scalar. Chain rule applies: $D_v f = \nabla f \cdot v$. Yes, the inner derivative can be described as $v$ (or $h$ in your case).
Same here. If $T$ has (complex) Gâteaux derivative $A$ at $\psi$, then by definition of this derivative we have $$T(\psi+zh) - T(\psi) = z A h +o(|z|),\quad z\to 0 \tag1$$ Here $A$ is a bounded linear operator from the function space into $l^1$. The $o(|z|)$ term is something that goes to zero norm-wise, even if divided by $|z|$).
After that $T$ is paired with $\lambda \in l^\infty$, i.e., a continuous linear functional is applied to it. Applying $\lambda$ to both sides of (1) we get $$\langle \lambda, T(\psi+zh) - T(\psi) \rangle = z \langle \lambda, A h \rangle +o(|z|),\quad z\to 0 \tag2$$ This $\langle \lambda, A h \rangle $ thing is the derivative.