Existence of Hermitian adjoint For an operator $F$, its (Hermitian) adjoint, $F^\dagger$, is defined as an operator satisfying
$$\int f^*F^\dagger g~\mathrm d^3r=\int(Ff)^*g~\mathrm d^3r,$$
for any arbitrary functions $f$, $g$ such that the integrals exist.
But what is the guarantee that such an operator, $F^\dagger$ with this property exists for any arbitrary operator $F\left(\mathbf r, \frac{1}{i}\nabla_{\mathbf r}\right)$?
I'm reading Merzbacher's Quantum Mechanics, and in it, he states that it can be verified by integration by parts. but I'm not sure how.
 A: The domain of $F^\dagger$ consists of all $g\in\mathcal H$ such that, for all $f\in Dom(F)$, the inner product $\left<g,Ff\right>$ can be written as $\left<h,f\right>$ for some $h\in \mathcal H$.  We then define $F^\dagger g := h$.

If $Dom(F)$ is dense in $\mathcal H$ then this $h$ is unique, for if $\left<h_1,f\right>=\left<h_2,f\right>$, then $\left<h_1-h_2,f\right> = 0$.  Since any $\psi\in \mathcal H$ can be written as the limit of a sequence $\{f_n\}$ (which is what it means for a subset to be dense), it follows that
$$\left<h_1-h_2,\psi\right>=\left<h_1-h_2,\lim_{n\rightarrow \infty}f_n\right>= \lim_{n\rightarrow \infty} \left<h_1-h_2,f_n\right> = 0$$
where we've used that the inner product is a continuous map.  Since this holds for all $\psi\in \mathcal H$, it follows that $h_1-h_2=0 \implies h_1 = h_2$.

The uniqueness of $h$ means that the operator $F^\dagger: g\mapsto h$ is well-defined on all $g\in Dom(F^\dagger)$. We also know that $Dom(F^\dagger)$ is non-empty, because no matter what $F$ is, $Dom(F^\dagger)\supseteq \{0\}$ (this is easy to check).
Therefore, it follows that any densely-defined operator $F$ has a well-defined adjoint $F^\dagger$.  In principle, it's possible that $F^\dagger$ is trivial, meaning that the only element of its domain is $0$.  On the other extreme, it may be that $F^\dagger = \mathcal H$, the entire Hilbert space.  Without additional information about $F$, it's not possible to be more specific about $F^\dagger$.

But what is the guarantee that such an operator, $F^\dagger$ with this property exists for any arbitrary operator $F(\mathbf r,\frac{1}{i}\nabla_\mathbf{r})$?

This is a subtle issue, and it's certainly not true that any operator which can be written as a "function" of the position and momentum operators is defined on a dense domain.  By way of example, the operator
$$(F\psi)(x)= \sum_{n=1}^\infty x^n\psi(x)$$
is well-defined only for $\psi$ which are zero outside the range $x\in(-1,1)$, which is certainly not a dense subset of $L^2(\mathbb R)$; this means that $F$ has no unique adjoint. However, if you restrict your attention to simple sums and products of $x$ and $p$, you can find the adjoint via integration by parts.
For example, consider the infinite square well, so $\mathcal H=L^2([0,1])$.  Define the operator $F = \frac{d}{dx}$ with domain $Dom(F) = C^1([0,1])$ - the continuously-differentiable functions on $[0,1]$.
We would then have
$$\left<g,Ff\right> = \int g(x)^* f'(x) dx \underbrace{ = }_{IBP} g(x)^*f(x)\bigg|^1_0 -\int g'(x)^* f(x) dx$$
In order for this to be of the form $\left<h,f\right>$ for some $h$, the boundary term needs to vanish, which means that we need $g(0)=g(1)=0$ (because $f$ can take any values at the endpoints).  In that case, we have$^\dagger$
$$F^\dagger = -\frac{d}{dx}$$
$$Dom(F^\dagger)= \left\{ g\in C^1([0,1]) \ \big| \ g(0)=g(1)=0\right\}$$

$^\dagger$This domain isn't quite right, but it's close enough for the present discussion.
