# What does it mean for a QFT to not be well-defined?

It is usually said that QED, for instance, is not a well-defined QFT. It has to be embedded or completed in order to make it consistent. Most of these arguments amount to using the renormalization group to extrapolate the coupling up to a huge scale, where it formally becomes infinite.

First of all, it seems bizarre to me to use the renormalization group to increase resolution (go to higher energies). The smearing in the RG procedure is irreversible, and the RG doesn't know about the non-universal parts of the theory which should be important at higher energies.

I have read in a note by Weinberg that Kallen was able to show that QED was sick using the spectral representation of the propagator. Essentially, he evaluated a small part of the spectral density and showed that it violated the inequalities for the field strength $Z$ that are imposed by unitarity + Lorentz invariance, but I do not have the reference.

So what I would like to know is :

What does it take to actually establish that a QFT is not well-defined?

In other words, if I write down some arbitrary Lagrangian, what test should I do to determine if the theory actually exists. Are asymptotically free theories the only well-defined ones?

My initial thought was: We would like to obtain QED (or some other ill-defined theory) using the RG on a deformation of some UV CFT. Perhaps there is no CFT with the correct operator content so that we obtain QED from the RG flow. However, I don't really know how to make that precise.

Any help is appreciated.