A UV complete theory is one whose correlation functions or amplitudes may be calculated and yield unambiguously finite results for arbitrarily high energies.
Yes, asymptotic freedom is enough for UV completeness because the UV limit of the theory is free and therefore well-defined. Whenever the coupling constant is small and the beta-function is negative at one-loop level, the higher-loop corrections should be small so that the exact beta-function should be negative, too. Such implications become even easier to make with SUSY.
Yes, exactly scale-invariant (and especially conformal) theories are UV complete if they're consistent at any scale because they predict the same thing at all scales due to the scale invariance. However, scale invariance at the leading order doesn't imply exact scale invariance. So no, one-loop vanishing of the beta-function may be coincidental and – even in a SUSY theory – the full beta-function may still have both signs.
Yes, superconformal theories are a subset of conformal theories so they're UV-complete. The S-duality exchanges descriptions with a different value of the coupling which is a different quantity, and therefore an independent operation/test, from the UV-completeness and scale invariance that relates different dimensionful scales.
It is believed that all conformal theories must have "some" Maldacena dual in the bulk although whether this dual obeys all the usual conditions of a "theory of quantum gravity" or even a "stringy vacuum" is unknown, especially because we can't even say what all these conditions are. The non-AdS/non-CFT correspondence would in principle work for all UV-complete theories but it's much less established and more phenomenological than the proper AdS/CFT correspondence that works with the exactly conformal theories only.
I) This is discussed around eq. (23.7.1) on p. 462 in Ref. 1. The task is to perform the path integral
$$\tag{1} \int_{BC} [d\phi]e^{\frac{i}{\hbar}S[\phi]} ~=~\sum_{\nu}\int\! du \int_{BC_0} [d\phi_q]e^{\frac{i}{\hbar}S[\phi_{cl}+\phi_{\nu,u}+\phi_q]}
$$
over fields $\phi$ with some (possible inhomogeneous) boundary conditions $BC$. This is done by splitting the fields
$$\tag{2} \phi~=~\phi_{cl}+\phi_{\nu,u}+\phi_{q}$$
into the following parts.
A single distinguished classical solution $\phi_{cl}$ (in the trivial instanton sector). The classical solution $\phi_{cl}$ satisfies the Euler-Lagrange equations with the (possible inhomogeneous) boundary conditions $BC$.
A set of instantons $\phi_{\nu,u}$ labelled with discrete topological number $\nu$ and continuous moduli $u$. The instantons $\phi_{\nu,u}$ satisfy the Euler-Lagrange equations with homogeneous boundary conditions $BC_0$. Instantons arise when there isn't a unique solution to the Euler-Lagrange equation with the given boundary conditions $BC$.
And quantum fluctuation $\phi_q$ satisfying the homogeneous boundary conditions $BC_0$.
The action
$$\tag{3} S[\phi_{\rm cl}+\phi_{\nu,u}+\phi_q]
~=~S[\phi_{\rm cl}+\phi_{\nu,u}]+S_{2}[\phi_q]+{\cal O}((\phi_q)^3)
~\approx~S[\phi_{\rm cl}+\phi_{\nu,u}]+S_{2}[\phi_q]$$
is then often expanded to quadratic order (denoted $S_2$) in the quantum fluctuations $\phi_q$ leading to a Gaussian path integral. See also eq. (23.7.2) in Ref. 1.
Note that the linear term $S_{1}[\phi_q]=0$ in $\phi_q$ vanishes because of Euler-Lagrange equations.
In ordinary perturbation theory without instantons, there is no summation over instanton sectors and integration over moduli.
II) One may wonder if the summation over instanton sectors in eq. (1) constitutes a kind of over-counting of the field configurations in the path integral? E.g. couldn't one reproduce a non-trivial instanton by including sufficiently many (all?) quantum corrections in the trivial sector, etc.?
From an idealized mathematical point of view, the need to sum over instanton sectors may be seen as the mathematical fact that not all $C^{\infty}$ functions are analytic. (Speaking of analycity, it seems relevant to mention the characteristic hallmark of instantons: The (non-trivial) instanton terms in the partition function have non-analytic dependence of the coupling constants of the theory. In short: One cannot reproduce non-perturbative effects by only applying perturbative methods.)
However in practice the path integral over quantum fluctuations is not well-defined (much) beyond the Gaussian approximation. So in practice one may view the decomposition on the rhs. of eq. (1) as a pragmatic definition of the full path integral on the lhs. of eq. (1).
References:
- S. Weinberg, The Quantum Theory of Fields, Vol. 2, p. 421 and p. 462.
Best Answer
I think both the link and the question refer to Dyson's heuristical argument on why the perturbative series in QED could not be convergent. It goes somewhat like this:
Suppose the series in $\alpha$ converges in some radius. The it converges also for negative values of the coupling constant inside that radius. Consider now what kind of theory is QED with a negative $\alpha$. In that theory like charges attract and opposite sign charges repel each other. Now take the vacuum of the non-interacting theory. This state is unstable against formation of electron-positron pairs, because said pairs would repel indefinitely leading t a lower energy state. You can make an even lower energy state by adding pairs that would separate in two clusters of electrons on one side and positrons on the other. Therefore this theory does not possess a ground state, since the spectrum is unbounded from below. Hence there is no consistent QED for negative coupling constant. And so the perturbative series cannot converge.
As far as I know this argument is strictly heuristical, but shortly after it appeared (in the 1950s) Walter Thirring proved the divergence for a particular scattering process (I'm not in my office so I don't have the correct reference, but I'm positive the paper is in Thirring's selected works as well as explained in his autobiography).
Note that this question of convergence was proeminent in a period where people tried to define QFT in terms of the perturbative expansion. The advent of non-perturbative effects (instantons, confinements, pick your favorite...) coupled with renormalization group showed that this was the wrong approach for QFT.
But note also that the argument of vacuum instability depends on the interaction. It does not preclude the possibility of designing a QFT with convergent perturbative expansion, it just shows that it it not to be expected in a general theory.