Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous periodic function. If we replace $f$ by $$g(t) := f(\frac{1}{\lambda}t)$$ for some $\lambda > 0$, this new function $g$ has period $\lambda$ times the period of $f$, and$\mathbf{{}^1}$ $$\hat{g}(s) = \lambda \hat{f}(\lambda s)$$.
That is $f$ has been pitch-shifted down by $\log_2 \lambda$ octaves, but also time-dilated by a factor of $\lambda$.
My question is: is there mathematically well-defined operation of pitch shift without time dilation?
I strongly suspect the answer is "no", because any continuous linear operation $T_{\lambda}$ which correctly pitch-shifts down by $\log_2 \lambda$ octaves all functions of the form $f(t) = sin(\omega t)$, i.e., for which $T_{\lambda}(sin( \omega t)) = sin(\frac{\omega}{\lambda} t)$, will necessarily have $T_{\lambda}(f(t)) = f(\frac{1}{\lambda}t)$ for every continuous periodic function $f$.
This doesn't immediately rule out a non-linear (or possibly non-continuous) operation, so I'm wondering if there is anything in the literature about such an operation, or a "no-go" theorem showing that the notion of pitch-shift without time-dilation is incoherent, period.
Most papers about this subject are written by engineers, who really only care about performing this operation approximately to give more-or-less the correct psycho-acoustic effect. E.g. this paper using wavelets, which I take to be roughly state-of-the-art. Other past approaches have been even worse: e.g. breaking up a signal into short time intervals doing a naive pitch-shift with time dilation (possibly repeating or clipping), and then assembling them back together, introducing discontinuities when moving from one time interval to the next, which are then smoothed over by some filter.
My motivation here is to be able to have an ideal standard by which to judge the degree of accuracy of a pitch-shift algorithm. If there is none, then the question of the accuracy of a pitch-shift algorithm is an empirical question about subjective experience rather than a mathematical one.
$\mathbf{{}^1}$ The Fourier transform will have to be a distribution rather than a function.