Post's structured addressing an originally different formulation of the question, "agree" and "disagree" responding to "can a bandlimited signal amplitude-alias?" - but still answers the current question.
Where I agree
If the actual, continuous/analog signal has no continuous Fourier transform frequency components above $f_s / 2$, then there can be no aliasing. This is explained in detail in Dan's answer.
Put differently, an $s(t)=A(t)\cos(\omega t)$ with amplitude $A(t)$ that samples to the observed waveform would necessarily have CFT components above $f_s/2$, whereas we assume there are none.
So why disagree?
Because Fourier isn't alpha and omega. I said the signal is bandlimited, or having a finite range of frequencies; I never said these were Fourier frequencies.
Consider a pure tone, $f=64$; now picture its Fourier transform (or DFT). Nice, clean, single spike at $\pm 64$. Now modulate its amplitude: $A(t) \cos(32 \pi t),\ A(t)=\cos(\pi t)$, sample at $f_s = 1290$, take DFT:
A mess! Nonzero frequencies far beyond $64$. For continuous FT, not only do we see a higher frequency, but the original $f=64$ is gone, and we have $f=\pm 63.5, \pm 64.5$ - neither of which are in the non-AM waveform. And DFT isn't the only one delicate; if $A(t)$ weren't purely sinusoidal, FT too would get messy.
Is there a nicer, more intuitive representation? Sure - EMD: "instead of constant amplitude and frequency in a simple harmonic component, an IMF can have variable frequency and amplitude along the time axis." So amplitude gets its own, independent representation, instead of squeezing it along frequency as constants. This enables vastly more intuitive and sensible descriptions of a fixed frequency that happens to be amplitude-modulated.
Does this matter in practice?
Absolutely. Consider a damped swinging pendulum, whose measurements we have (assume without noise). Don't assume constant damping; just assume it never changes the cycle periods (left & right stop times). At what frequency is the pendulum swinging? Suppose it's actually 2 / sec, but we don't know that.
There's no way to spin the story or measure differently that wouldn't result in the Fourier interpretation telling us "it's 2 and-some!" And the "some" can be 4, 8, 40, or more - an entire nonzero distribution of these.
Does this make much sense? Do we get an accurate representation of the physical process? Absolutely not; there are no $f=8, 40$, and the other frequencies. These are what's demanded by Fourier bases assuming fixed amplitudes, and is precisely what this warmly-received answer further illustrates.
With an EMD lens, however, we get a much cleaner and sensible representation: $A(t) \cos (4\pi t)$, with the one frequency of the actual source process being clearly captured, and amplitude can now be analyzed independently for various factors.
Let me repeat: there are no higher (or lower) frequencies in the physical process, and $\mathcal{F}$ will strongly beg to differ.
My own application is what motivated the question to begin with.
The goal is to preprocess EEG data as a "feature extraction" step before feeding it to a neural network for seizure classification. As such, it's critical I meaningfully interpret the interactions of used transforms with data.
Below are two different Continuous Wavelet Transform implementations applied to the same signal:
One clearly represents the right part's dip in amplitude, other doesn't. The waveform is two adjacent signals, $f=64, 1$, each with $N=129$ samples. So the actual, continuous / analog waveform is more accurately captured by one on the right. ... but how do I know this, in practice?
Of course if I know the data is Fourier-bandlimited, I can state with certainty that this is a 1Hz pure tone with an "imaging" effect, rather than an AM waveform. But I cannot possibly know that; Fourier frequencies aren't even 'actual' frequencies to begin with, as previously noted, so the only way to know if a signal is Fourier-bandlimited is to either have the source's mathematical function (good luck), or make a series of assumptions, apply filters, and be "sufficiently confident" the assumption holds.
What I can know is how bandlimited the physcal process is. For EEG, we're dealing with the brain, which we know to be capable of producing waves under a certain frequency (mostly <100Hz IIRC). If we strap on a device that samples at 10kHz, we'll undoubtedly get a DFT with nonzero normalized frequency components in thousands - which is absurd.
Thus we have a piece of information that's incompatible with a purely Fourier lens. If only we could avoid such 'artificial frequencies' by allowing amplitude to be represented independently... hello EMD.
Here's the truth: you either cannot tell me the physical process's, like one's this waveform may have emerged from, Fourier $f_\text{max}$, or it may cost you a lot to figure out. But you can tell me, from knowledge of the system, its physical, 'actual' $f_\text{max}$, which is precisely mathematically captured by decoupling frequency and amplitude representations as in EMD. If it's "EMD-bandlimited", it can very well be amplitude aliasing, and we can treat it accordingly.
What this means for responders & why it matters, in Meta.
Mathematical formulation of "bandlimited" wasn't the purpose of this question, but it's a valid inquiry of its own. Doubt I'll dedicate a Q&A to it so here briefly:
One must restrict what qualifies as "amplitude", $A(t)$, else it could be anything, including something that divides out whatever it modulates, making the definition useless. An intuitive restriction, aside continuity and differentiability, is $\mathcal{F}(A(t)) \ll \mathcal{F}e^{j\phi(t)}$ for
$$
s(t) = \sum_k A_k(t) e^{j\phi(t)}
$$
Then "bandlimited" would be defined over $\omega_k$:
$$
\omega_k (t) = \frac{d}{dt} \phi_k (t)
$$
Such construction is applied in Daubechies et al - page 538:
"Bandwidth" is defined over $\omega_k$ above - thus, so is bandlimited. A more rigorous formulation is given in this paper:
Further explorations and applications in Lilly & Olhede and much other literature.
