Let
- $U_0,H$ be separable $\mathbb R$-Hilbert spaces
- $\mathfrak L(U_0,H)$ denote the space of bounded linear operators from $U_0$ to $H$
- $\text{HS}(U_0,H)$ be the set of all $Q\in\mathfrak L(U_0,H)$ with $$\sum_{n\in\mathbb N}\left\|Qe_0^n\right\|_H^2<\infty\tag 1$$ for all orthonormal bases $(e_0^n)_{n\in\mathbb N}$ of $U_0$
- $G:H\to\operatorname{HS}(U_0,H)$ with $G\in C^1(H,\mathfrak L(U_0,H))$
Now, let $(e_0^n)_{n\in\mathbb N}$ be an orthonormal basis of $U_0$ and $$w^N(x):=\sum_{n=1}^N\left({\rm D}G(x)G(x)e_0^n\right)e_0^n\;\;\;\text{for }x\in H$$ for $N\in\mathbb N$.
Can we show that $(w^N(x))_{N\in\mathbb N}$ is convergent in $H$ for all $x\in H$?
I guess not, cause we can only show that
\begin{equation} \begin{split} \left\|w^N(x)\right\|_H&\stackrel{\text{triangle inequality}}\le\sum_{n=1}^N\left\|\left({\rm D}G(x)G(x)e_0^n\right)e_0^n\right\|_H\\ &\stackrel{\text{definition of }\left\|\;\cdot\;\right\|_{\mathfrak L(U_0,\:H)}}\le\sum_{n=1}^N\left\|{\rm D}G(x)G(x)e_0^n\right\|_{\mathfrak L(U_0,\:H)}\underbrace{\left\|e_0^n\right\|_{U_0}}_{=\:1}\\ &\stackrel{\text{definition of }\left\|\;\cdot\;\right\|_{\mathfrak L(H,\mathfrak \:L(U_0,\:H))}}\le\left\|{\rm D}G(x)\right\|_{\mathfrak L(H,\mathfrak \:L(U_0,\:H))}\sum_{n=1}^N\left\|G(x)e_0^n\right\|_H \end{split}\tag 2 \end{equation}
for all $N\in\mathbb N$ and $x\in H$ and we only know that $\left(\sum_{n=1}^N\left\|G(x)e_0^n\right\|_H^{\color{red}{2}}\right)_{N\in\mathbb N}$ is convergent for all $x\in H$ by assumption.
So, if my conclusion isn't wrong, which additional assumption do we need?