Problem
Show that for any two operators $\hat{A}$ and $\hat{B}$, the adjoint $(\hat{A} + \hat{B})^\dagger = \hat{A}^\dagger + \hat{B}^\dagger$. Do so using the integral form of the definition of adjoint.
I know how to show this by manipulating the definition of the adjoint written in bra-ket notation, however I'm unable to show the same using the "integral form". Specifically, we're expected to do the following:
By the definition of adjoint:
$$\langle f | (\hat{A} + \hat{B})^\dagger g\rangle \equiv \langle (\hat{A} + \hat{B})f | g \rangle$$
...where $\dagger$ denotes the adjoint, and $\star$ denotes the complex conjugate
This may be re-expressed as:
$$\int f^\star (\hat{A} + \hat{B})^\dagger g = \int \bigg[(\hat{A} + \hat{B}) f\bigg]^\star g$$
We are then asked to re-express the RHS of the equation above such that it is of the form of the LHS. Distributing the complex conjugation, I get:
$$\int f^\star (\hat{A} + \hat{B})^\dagger g= \int (\hat{A} + \hat{B})^\star f^\star g = \int (\hat{A}^\star + \hat{B}^\star)^\star g$$
To get this in the form of the LHS, we must move $f^\star$ to the left of $(\hat{A}^\dagger + \hat{B}^\dagger)$. Unfortunately, it isn't clear to me how this can be done.
Sorry, this is a mathematics site, formal manipulation in physicist's notation will simply not cut it.
To do this properly, you need to talk about domains of these operators, which in general are not bounded. $A^\dagger$ can only be defined (by $\langle x, A^\dagger y\rangle = \langle Ax, y \rangle$ for $x \in \mathcal D(A)$) if the domain $\mathcal D(A)$ is dense in your Hilbert space. Now the natural definition of the domain of $A+B$ is $\mathcal D(A) \cap \mathcal D(B)$. Unfortunately, in general there is no reason to think that $\mathcal D(A) \cap \mathcal D(B)$ is dense even though $\mathcal D(A)$ and $\mathcal D(B)$ are. In fact, this intersection may well contain no nonzero vectors at all. Thus as a general statement $(A+B)^\dagger = A^\dagger + B^\dagger$ does not make any sense.
For example, let $H$ be a densely-defined self-adjoint operator on a separable infinite-dimensional Hilbert space $\mathcal H$ with purely discrete spectrum, and $U$ a unitary operator on $\mathcal H$. Then for a "generic" $U$ in the sense of Baire category (i.e. a member of the intersection of countably many open dense subsets of the space of unitary operators on $\mathcal H$, $\mathcal D(H) \cap \mathcal D(U H U^\dagger) = \{0\}$. See my paper Some Generic Results in Mathematical Physics, Markov Processes and Related Fields 10 (2004), 517-521.
EDIT: OK, so let's just assume $\mathcal D(A+B) = \mathcal D(A) \cap \mathcal D(B)$ is dense in the Hilbert space $\mathcal H$. The domain $\mathcal D(A^\dagger)$ is defined as the set of $y \in \mathcal H$ such that the functional $x \mapsto \langle Ax, y \rangle$ on $\mathcal D(A)$ is bounded. In that case there is a unique $z \in \mathcal H$ such that $\langle Ax, y \rangle = \langle x, z\rangle$, and we define $A^\dagger y$ to be this $z$. Similarly for $B^\dagger$ and $(A+B)^\dagger$.
Then it is true that if $y \in \mathcal D(A^\dagger) \cap \mathcal D(B^\dagger)$, then $$\langle (A+B)x, y \rangle = \langle Ax, y \rangle +\langle Bx, y \rangle = \langle x, A^\dagger y \rangle + \langle x, B^\dagger y \rangle = \langle x, A^\dagger y + B^\dagger y \rangle$$ so $y \in \mathcal D((A+B)^\dagger)$ and $(A+B)^\dagger y = A^\dagger y + B^\dagger y$. Unfortunately we can't go in the other direction: there is no reason to expect every member of $\mathcal D((A+B)^\dagger)$ to be in $\mathcal D(A^\dagger)$ or $\mathcal D(B^\dagger)$. For example, we might have $B = -A$, in which case $(A+B)^\dagger = 0$ with domain all of $\mathcal H$. So it's still incorrect to say $(A+B)^\dagger = A^\dagger + B^\dagger$, just $A^\dagger + B^\dagger \subseteq (A+B)^\dagger$.