Bounded and Unbounded Operator Can someone explain with a concrete example of how can I can check whether a quantum mechanical operator is bounded or unbounded?
EDIT: For example., I would like to check whether $\hat p=-i\hbar\frac{\partial}{\partial x}$ is bounded or not. 
 A: A linear operator $A: D(A) \to {\cal H}$ with $D(A) \subset {\cal H}$ a subspace and ${\cal H}$ a Hilbert space (a normed space could be enough), is said to be bounded if:
$$\sup_{\psi \in D(A)\:, ||\psi|| \neq 0} \frac{||A\psi||}{||\psi||} < +\infty\:.$$
In this case the LHS is indicated by $||A||$ and it is called the norm of $A$.
Notice that, therefore, boundedness, is not referred to the set of values $A\psi$, which is always unbounded if $A\neq 0$, as $||A\lambda\psi|| = |\lambda|\: ||A\psi||$ for $\psi \in D(A)$ and $\lambda$ can be chosen arbitrarily large still satisfying $\lambda \psi \in D(A)$ since $D(A)$ is an subspace.
It is possible to prove that  $A: D(A) \to {\cal H}$  is bounded if and only if, for every $\psi_0 \in D(A)$:
$$\lim_{\psi \to \psi_0} A\psi = A \psi_0\:.$$ 
Another remarkable result is that a self-adjoint operator is bounded if and only if its domain is the whole Hilbert space. 
Regarding $A= \frac{d}{dx}$, first of all you should define its domain to discuss boundedness. An important domain is the space ${\cal S}(\mathbb R)$ of Schwartz functions since, if $-id/dx$ is defined thereon, it turns out Hermitian  and it admits only one self-adjoint extension that it is nothing but the momentum operator.  
$d/dx$ on ${\cal S}(\mathbb R)$ is unbounded. The shortest way to prove it is passing to Fourier transform. 
Fourier transform is unitary so that it transforms (un)bounded operators into (un)bounded operators. ${\cal S}(\mathbb R)$ is invariant under Fourier transform, and $d/dx$ is transformed to the multiplicative operator $ik$ I henceforth   denote by $\hat A$. So we end up with studying boundedness of the operator:
$$(\hat A \hat{\psi})(k) = ik \hat{\psi}(k)\:,\quad \hat\psi \in {\cal S}(\mathbb R)\:. $$
Fix $\hat\psi_0 \in  {\cal S}(\mathbb R)$ with $||\hat\psi_0||=1$ assuming that $\hat\psi_{0}$ vanishes outside $[0,1]$ (there is always such function as $C_0^\infty(\mathbb R) \subset {\cal S}(\mathbb R)$ and there is a function of the first space supported in every compact set in $\mathbb R$), and consider the class of functions 
$$\hat\psi_n(k):= \hat \psi_{0}(k- n)$$
Obviously, $\hat\psi_n \in {\cal S}(\mathbb R)$ and translational invariance of the integral implies $||\hat\psi_n||=||\hat\psi_0||=1$. Next, notice that:
$$\frac{||\hat A\hat\psi_n||^2}{||\hat\psi_n||^2} = \int_{[n, n+1]} |x|^2 |\hat\psi_{0}(k-n)|^2 dk \geq \int_{[n, n+1]} n^2  |\hat\psi_{0}(k-n)|^2 dk$$
$$ =
n^2 \int_{[0,1]} |\hat\psi_{0}(k)|^2 dk  = n^2\:.$$
We conclude that:
$$\sup_{\hat{\psi} \in {\cal S}(\mathbb R)\:, ||\hat{\psi}||\neq 0} \frac{||\hat A\hat\psi||}{||\hat\psi||} \geq n \quad \forall n\in \mathbb N $$
So $\hat A$ is unbounded and $A$ is consequently.
