It's good that you're considering questions like this; I find that this type of questions really forces a student to a deeper understanding of the math involved.
Do we then need any boundary conditions?
Yes, boundary conditions should be considered as part of the definition of the Hamiltonian and its domain. Different boundary conditions can result in different eigenfunctions/eigenvalues.
I somehow feel as if these Born von Karmann boundary conditions are not necessary in the sense that any eigenfunction to this <PERSON> operator is automatically periodic with the potential's period, is this true?
No, even with BvK conditions, this is not true (see <PERSON> answer for examples). What is true is that (with BvK), the wavefunction is periodic up to an an $e^{ikx}$ phase. Without BvK, neither of those statements is necessarily true.
In that case: Why do we want Born von Karmann boundary conditions?- My problem with the Born von Karmann conditions is that I find that they are not really boundary conditions, as they don't act on some boundary.
A couple things:
1. The hope in solid-state physics is that, if you have a big enough region, and you are computing bulk properties, then the edges form a tiny part of your system compared to the volume, and so the results should not depend much on which boundary conditions you choose.
2. BvK are boundary conditions: they say that the wavefunction at the left boundary of the space is the same as the wavefunction at the right boundary of the space.
3. BvK are useful, because they keep the discrete translational symmetry of the potential [translate by $a$ and nothing happens]. This is the symmetry which the Bloch waves represent, and if you choose boundary conditions which do not preserve that symmetry, then you will not have solutions which reflect that symmetry.
4. If you took some other set of boundary conditions, eg wavefunction goes to zero at both infinities, then your eigenfunctions (if they exist) would not be Bloch waves. They might be able to look a lot like Bloch waves over some finite region, but would have to fall off at some point. By definition, Bloch waves have the same norm in every unit cell.
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BvK provides you with conveniently normalizable eigenfunctions, since your space is finite. You don't have to worry about integrating out to infinity for the unbound states and having the norm explode.
In my physics lecture we noticed that due to these <PERSON> conditions the possible k′s for the problem (appearing in the exponentials) are discrete. Not sure if this is automatically satisfied, even if we don't assume Born von Karmann boundary conditions? True, this is not a special thing about BvK. If you designed the problem so the periodic lattice actually was large but finite and took the boundary conditions to go to zero at infinity, then you would find a discrete spectrum of bound states in the lattice region that look something like <PERSON> states deep in the lattice but then fall off at the edges. (I assume the lattice potential is strong enough to have bound states). This is, of course, a more realistic case then BvK, since materials are finite in the real world...but you don't do it because solving for eigenfunctions just became quite ridiculous.
Then we said that for every k, the <PERSON> operator equation that you get by pluggin in the ansatz from the <PERSON> or <PERSON> thoerem has a discrete spectrum. Is this true? If so, what does this all have to do with bands, if everything is so nicely discrete? - Or do we just cal these things bands, since the k's get so close, that we cannot really resolve the discrete structure?
Exactly. If we have well-separated nondegenerate bands, then lowest "band" is the set of {for each k, take the lowest eigenvalue}. The second lowest band is then {for each k, take the second lowest eigenvalue}. Etc. Since the allowed k's are so close as to basically be unresolvable, each band looks like a continuum of states. 795