Let $H$ be a Hilbert space, $I$ a set and $(A_i)_{i\in I}\in\mathcal{B}(H)^I$ a family of bounded operators. In Halmos' Introduction to Hilbert Space and the Theory of Spectral Multiplicity the author defines the family $(A_i)$ to be summable with sum $A\in\mathcal{B}(H)$ if for every $f\in H$ the family $A_i f$ is summable in the sense that for every $\epsilon >0$ there exists a finite set $J\subset I$ such that for every finite set $J\subset K\subset I$, $$\Big\|Af-\sum_{k\in K}A_kf\Big\|\leq\epsilon$$ In notation: $A=\sum_i A_i$.
Question 1. Suppose $(A_i)_i$ is summable with sum $A$. Does it follow that $(A_i^*)$ is summable?
Here $T^*$ denotes the adjoint operator of $T$ characterized by $\forall f,g\in H, \langle Tx\mid y\rangle = \langle x\mid T^*y\rangle$. I'm not sure I believe the answer to be positive. If it isn't, are there simple counter examples ?
Question 2 (not really a question). Suppose both $(A_i)_i$ and $(A_i^*)_i$ are summable with sums $A$ and $B$. Does it follow that $A^*=B$ i.e. do we have $(\sum_i A_i)^* = \sum_i A_i^*$?
This question has a simple affirmative answer: if $f,g\in H$ then using standard properties of summable families of vectors \begin{align} \langle Af\mid g\rangle &=\langle \sum_i A_i f\mid g\rangle\\ &=\sum_i \langle A_i f\mid g\rangle\\ &=\sum_i \langle f\mid A_i^* g\rangle\\ &=\langle f\mid \sum_i A_i^* g\rangle\\ &=\langle f\mid B g\rangle \end{align}
Question 3. If the answer to question 1 is indeed negative, under what stronger conditions does the conclusion hold?
The text provides a very partial answer to this question when the $A_i$ are assumed to be (orthogonal) projections (onto closed subspaces) and $A$ is a projection. Since the goal is to define integrals wrt projection valued measures I can see how there might be no incentive to pursue this question any further. I ask out of pure curiosity.
Let $f\in H$: to prove summability of the family $A_i^*f$ one would need to show that $$\forall\epsilon>0\exists J\in\mathcal{P}_{\mathrm{fin}}(I)\forall K'\in \mathcal{P}_{\mathrm{fin}}(I), \Big(J\sqcup K'\implies\Big\|\sum_{k'\in K'}A_{k'}^*f\Big\|\leq\epsilon\Big)$$ One may then try to write, for $K'$ finite subset of $I$ and $A_{K'}^*:=\sum_{k'\in K'}A_{k'}^*$, $$\Big\|\sum_{k'\in K'}A_{k'}^*f\Big\|=\|A_{K'}^*f\|=\sup_{\|g\|=1}|\langle g\mid A_{K'}^*f\rangle|=\sup_{\|g\|=1}|\langle A_{K'}g\mid f\rangle|$$ Choose a $g$ that comes close to realizing this supremum. But then one would like to have flexibility for the set $K'$ which however doesn't play nicely with the fact that $g$ was chosen with a particular $K'$ in mind.