What is the etymology of the term "wave function"? Question: If the wave function $\psi \to 0$ (pointwise or uniformly in mean), then does $|\psi|^2$ converge in quadratic mean (i.e. in $L^2$) to $0$?
Is this the source of the term "wave function"? 

(Since the energy of a wave is proportional to the square of its amplitude, so if the expected energy of a wave goes to zero, then its ensemble amplitude goes to zero in quadratic mean. Likewise, if the expectation of a wave function goes to zero, then its probability density will decay to zero in $L^2$/quadratic mean. And the wave function is supposed to have something to do with energy, hence why it being square-integrable has something to do with it having finite energy, and thus in a certain sense is related to an elementary wave.)

Note: This is a follow-up to my previous question: Do waves and oscillations dissipating energy decrease in amplitude in mean square?
 A: The connection between wavefunctions and waves is much more fundamental than that. The quantum wavefunction $\psi(x)$ has a huge amount in common with the amplitude of a classical wave. For concreteness, let's consider the height $y(x)$ of a wave on a string at some time $t$.


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*The equations of motion for $\psi(x)$ and $y(x)$ are both wave equations, i.e. they both admit traveling wave solutions $e^{i(kx-\omega(k) t)}$. The only difference is the dispersion relation $\omega(k)$.

*The equations of motion for both types of waves are linear. This implies that wavefunctions, as well as string waves, can display constructive or destructive interference.

*Given fixed boundary conditions, both $\psi(x)$ and $y(x)$ have standing wave solutions with quantized frequencies/energies $\omega$.

*The density of 'stuff' in the wave is proportional to $|\psi(x)|^2$ (probability) and to $y(x)^2$ (energy). 


This last point is what you found, just stated in simpler words. Physical wave(functions) must be $L^2$ functions, as probability must be normalized and energy must be finite, respectively.
