Self-adjoint and unbounded operators in QM 
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*An operator $A$ is said to be self-adjoint if $(\chi,A\psi)=(A\chi,\psi)$ for $\psi, \chi \in D_A$ and $D_A=D_{A^\dagger}$. But for the free particle momentum operator $\hat{p}$ these inner products does not exist, however its eigenvalues are real. So, is $\hat{p}$ a self-adjoint operator? 

*Why are the operators in quantum mechanics in general unbounded?
 A: If $\cal H$ is a complex Hilbert space, and  $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a unique operator, the adjoint $A^\dagger$ of $A$ satisfying (this is its definition)
$$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad 
\forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$
 with:
$$D(A^\dagger) := \{ \phi \in {\cal H}\:|\: \exists  \phi_1 \in {\cal H} \mbox{ with}  \: \langle \phi_1 |\psi \rangle = \langle \phi | A \psi\rangle \:\: \forall \psi \in D(A)\}$$
The above densely-defined operator $A$ is said to be self-adjoint if $A= A^\dagger$. A densely-defined  operator satisfying
$$\langle A \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \psi,\phi \in D(A)$$
is said to be symmetric. It is clear that the adjoint of $A$, in this case, is an extension of $A$ itself. 
A symmetric operator is essentially selfadjoint if $A^\dagger$ is self-adjoint. It is possible to prove that it is equivalent to say that $A$ admits a unique self-adjoint extension (given by $(A^\dagger)^\dagger$).
The operator $-i\frac{d}{dx}$ with domain given by  Schwartz' space ${\cal S}(\mathbb R)$ (but everything follows holds also if the initial domain is $C_0^\infty(\mathbb R)$) is symmetric and essentially self-adjoint. Both $-i\frac{d}{dx}$ and the true momentum operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$ are not bounded. 
Both operators do not admit eigenvalues and eigenvectors. 
The spectrum of $p$ is continuous and coincides with the whole real line.
Passing to Fourier-Plancherel transform, the operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$ turns out to coincide with the multiplicative operator $k \cdot$.
Concerning the issue of unboundedness of most self-adjoint quantum operators, the point is that a celebrated theorem (one of the possible versions of Hellinger–Toeplitz theorem) establishes that:
a (densely-defined) self-adjoint operator 
$A :D(A) \to \cal H$ is bounded if and only if $D(A)= \cal H$
and almost all 
operators of QM, for various reasons, are not defined in the whole Hilbert space (unless the space is finite dimensional). These operators are not initially defined on the whole Hilbert space because they usually are differential operators. Differential operators need some degree of regularity to be applied on a function, whereas the generic element of a $L^2$ space is incredibly non-regular (it is defined up to zero-measure sets). The subsequent extension to self-adjoint operators exploits a weaker notion of derivative (weak derivative in the sense of Sobolev) but the so-obtained larger  domain is however very small with respect to the whole $L^2$ space.
ADDENDUM. In view of a remarkable Andreas Blass' comment, I think it is worth stressing a further physical reason for unboundedness of some self-adjoint operators representing observables in QM.
First of all the spectrum $\sigma(A)$ of a self-adjoint observable represented by a self-adjoint operator $A$ has the physical meaning of the set of all possible values of the observable. So if the observable takes an unbounded set of values, the spectrum $\sigma(A)$ must be an unbounded subset of $\mathbb R$.
Secondly, if $A$ is a self-adjoint operator (more generally a normal operator), the important result holds true:
$$||A|| = \sup\{ |\lambda| \:|\: \lambda \in \sigma(A)\}$$
including the unbounded cases where both sides are $+\infty$ simultaneously.
So, if an observable, like $p$ or $x$ or the angular momentum,  takes an unbounded set of values, the self-adjoint representing it has necessarily to be unbounded (and therefore defined in a proper dense subspace of the Hilbert space).
