Questions regarding $D=4 $ ${cal N}=4$ supersymmetric Yang-Mills

Question

I have some questions regarding the $D=4 $ ${cal N}=4$ super-Yang-Mills theory (the one with a really long action which can be acquired by compactifying the 10-dimensional ${cal N}=1$ theory).

I have heard that the theory has a vanishing beta-function, so the superconformal symmetry is not anomalous. Because there are beta-functions involved, I am obligated to ask whether this is a perturbative result. I am almost sure that it isn't, but still.

If it is exact, how can I show that? I am not asking for the absolutely rigorous proof, but I need something better then "Feynman diagrams vanish", or at least why can we treat this non-perturbative problem perturbatively.
Another (related) question: is this theory exactly solvable? If so, how does one recover the correlations? (My assumption would be that some kind of a bootstrap method can be used, since the superconformal symmetry is exact).
And finally, does it mean that all anomalous dimensions (of fields and couplings) are zero (meaning that the classical RG flow completely determines how the theory is influenced by rescalings)? Is it correct to say that, unlike the usual Yang-Mills theory, N=4 SYM shows no fractal behaviour?

P.S. I am adding a string-theory tag despite the fact that this question has little to do with string theory, because I would expect string theorists to know a lot about SYM because of the AdS/CFT. It is also the reason I am studying it.

Nogueira · Accepted Answer

There is a non-perturbative proof that the theory is a CFT. See this paper. The idea of that paper is to use the fact that N=2 supersymmetry restrict the low energy effective action to be analytic in the N=2 prepotential and that the prepotential have good properties under $U(1)_{R}$ symmetry.
About your second question I can only refer to you this review.
The action does not receive anomalous corrections since the theory is a CFT. The theory has only one coupling, and this coupling sits in front of the action so it does not receive anomalous corrections to the dimension as well. The fields in general will receive anomalous corrections to the dimension, only some special family of operators that does not receive corrections such as the BPS operators that are protected by supersymmetry. The Lagrangian of the theory is one of these BPS operators (this is another way to see why the theory is a CFT).

Questions regarding $D=4 $ ${cal N}=4$ supersymmetric Yang-Mills

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