This is a rather subtle question as also at classical level the parameters of the theory could turn out to be redefined in a finite way. But, as far as you may know, we are very good with free theories and these mostly happen as fixed points for known theories. If you want an example you can take a look to the scalar field theory. You can consider a standard action like

$$S=\int d^4x\left[\frac{1}{2}(\partial\phi)^2-\frac{\lambda}{4}\phi^4\right]$$

This theory is trivial and this means that it reaches a trivial fixed point both at ultraviolet and infrared that makes it useless to describe physics unless some cut-off is introduced explicitly somewhere. But in the infrared you will get a beta function going like

$$\beta(\lambda)=4\lambda+\frac{c_1}{\sqrt{\lambda}}+O(1/\lambda)$$

and so, if you have a starting coupling $\lambda=\lambda_0$ you will get a running coupling going to zero like $p^4$ lowering momenta. The theory becomes free but these free excitations are all massive with a mass proportional to $\lambda_0^\frac{1}{4}$ as can also be seen from lattice computations. You can see from this that, notwithstanding we are coping with a trivial fixed point, all the parameters of the theory turn out to be properly redefined and in a finite way!

The meaning of this is that renormalization just expresses a physical property of a quantum theory: The simple fact that interaction changes all the parameters of a given theory when the couplings are turned on. But a trace of this can be found in the fixed points of the theory itself.

Now, if you look at the classical theory, you will be able to solve it exactly but the solutions have not finite energy unless you work with a finite volume or redefine the coupling $\lambda$, exactly as happens to the quantum theory. Also in this case you will get a mass even if you started from a massless theory and your coupling will run.

Again, we see that the effect of the interaction, the fact that the coupling $\lambda$ is not zero, is exactly to modify all the parameters of a theory.

As you see, this is true independently from the fact that you are coping with infinities or not.

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