Example of frames for Sobolev spaces A sequence of distinct vectors $\{f_1,f_2,...\}$ belonging to a separable Hilbert space $H$ is said to be a Frame if there exist positive contants $A$, $B$ such that, for $A<B$ and for all $f\in H$:
$$A\|f\|^2\leq\sum_{n=1}^\infty |(f,f_n)|^2\leq B \|f\|^2, \ \ \ (\ast)$$
An operator $S$ on $H$ is said "frame operator" of frame $\{f_1,f_2,...\}$, if
$$Sf=\sum_{n=1}^\infty (f,f_n) f_n$$
By this the $(\ast)$ becomes,
$$A\|f\|^2\leq (Sf,f)\leq B \|f\|^2$$
This is the definition for abstract frames. My question is: Are there any examples of frames for Sobolev spaces in the literature?
 A: The only context I know these frames from are Gabor frames. In that case, you are not interested to find an orthonormal basis, that has been done before, but rather find a "nice" (in that context: easy to generate) set of  vectors $\{g_i \}_{i \in I}$  such that
$$
f \approx \sum_{i=1}^Na_ig_i
$$
for all $f \in H$.
Of course, we could take the eigenfunctions of the Laplacian, but how about we consider starting with a function $\phi \in H^1(\mathbb{R^n})$ (maybe a Gaussian) and consider some set of discrete set of translations/modulations. Something like
$$
\phi_{(\omega,t)}=e^{-i\omega \cdot x}\phi(x+t)
$$
for some set $I$ such that $(\omega,t)\in I$. This set might for example be generated by some lattice.
If you impose further restrictions on $f$, which lead to a subspace of $H$, you can maybe choose your starting function more carefully. If I remember correctly, you almost always start with a Gaussian.
Now, just google Gabor frames for Sobolev spaces and you will find a few results, which are not just an orthonormal basis.
However, since it is not my exact field of expertise, please do not take my word as gospel and maybe some one else can recommend something specific. For now I will leave this answer up.
A: Based on your definition, a trivial example of a frame seems to be any (countable) orthonormal system  $\{f_k\}_{k \in \mathbb{N}}$ which exists for every separable Hilbert space $H$. For this choice of frame, the first inequality is satisfied with $A=1$ and $B=2$. The frame operator $S$ is just the identity. This would also work for a Sobolev space, of course.
