quantum mechanics - Considering an arbitrary wavefunction for a free particle, are all normalizable functions valid?

One can show that a possible solution to a wavefunction with constantly zero potential is equal to, only considering the spacial piece:

$$\psi(x) = \int_{-\infty}^{\infty} A(k) \ e^{ikx} dk$$

This is partially due to the fact that, since there are no boundary conditions applying a rule to $k$, such as $k = \frac{n \pi}{L}$ like in the case of the infinite square well. That merely explains why a sum of eigenstates cannot represent an arbitrary ket vector. This is, however, mainly due to the fact that a sum is not only unnecessary but inoperable, as a Fourier sum without boundary conditions like the infinite square well will be periodic throughout space and hence will not be normalizable and have weird implications on its position and momentum uncertainty I would think. This is why a Fourier integral is necessary.

However, if I am to postulate that $A(k)$ is the Fourier transform of $\psi(x)$, then could I not contend that this is just the Fourier transform of some arbitrary function $f(x)$ in disguise, and the solution to Schrodinger's equation given constant zero potential? In this case, since $f(x)$ is arbitrary, any function would be a solution given this case, as long as it is normalizable. Why or why not?