Is there a cousin of the Fourier transform which obeys the following property: if I rescale and translate a function, then its transformed form, at each frequency, only depends on the original transformed function at that frequency? In detail:

Given a function $f$, we can take its Fourier transform to get $\mathcal{F}(f)$:

$\mathcal{F}(f)(\omega)=\int e^{-2\pi ix \omega}f(x) dx$

If we translate $f$ by $\delta$ to get $f'(x)=f(x-\delta)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=e^{-2\pi i\delta \omega}\mathcal{F}(f)(\omega)$. In other words, $\mathcal{F}(f')(\omega)$ is a function of $\mathcal{F}(f)(\omega)$, $\delta$, and $\omega$.

If we scale $f$ by $a$ to get $f'(x)=f(x/a)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=\frac{1}{|a|}\mathcal{F}(f)(\omega/a)$. In other words, $\mathcal{F}(f')(\omega)$ is no longer a function of $\mathcal{F}(f)(\omega)$, but instead is a function of $\mathcal{F}(f)(\omega/a)$.

Does there exist a hypothetical new transformation $\mathcal{T}$, like the Fourier transform, but which obeys the following property: If $f'(x)=f(ax-\delta)$, then $\mathcal{T}(f')(\omega)$ is a function only of $\mathcal{T}(f)(\omega)$, $\omega$, $a$, and $\delta$.

Is there a cousin of the Fourier transform which obeys the following property: if I rescale and translate a function, then its transformed form, at each frequency, only depends on the original transformed function at that frequency? In detail:

Given a function $f$, we can take its Fourier transform to get $\mathcal{F}(f)$:

$\mathcal{F}(f)(\omega)=\int e^{-2\pi ix \omega}f(x) dx$

If we translate $f$ by $\delta$ to get $g(x)=f(x-\delta)$, the Fourier transform of $g$ obeys $\mathcal{F}(g)(\omega)=e^{-2\pi i\delta \omega}\mathcal{F}(f)(\omega)$. In other words, $\mathcal{F}(g)(\omega)$ is a function of $\mathcal{F}(f)(\omega)$, $\delta$, and $\omega$.

If we scale $f$ by $a$ to get $g(x)=f(x/a)$, the Fourier transform of $g$ obeys $\mathcal{F}(g)(\omega)=\frac{1}{|a|}\mathcal{F}(f)(\omega/a)$. In other words, $\mathcal{F}(g)(\omega)$ is no longer a function of $\mathcal{F}(f)(\omega)$, but instead is a function of $\mathcal{F}(f)(\omega/a)$.

Does there exist a hypothetical new transformation $\mathcal{T}$, like the Fourier transform, but which obeys the following property: If $g(x)=f(ax-\delta)$, then $\mathcal{T}(g)(\omega)$ is a function only of $\mathcal{T}(f)(\omega)$, $\omega$, $a$, and $\delta$.

Is there a cousin of the Fourier transform which obeys the following property: if I rescale and translate a function, then its transformed form, at each frequency, only depends on the original transformed function at that frequency? In detail:

Given a function $f:R\to R$, we can take its Fourier transform to get $\mathcal{F}(f):R\to R$:

$\mathcal{F}(f)(\omega)=\int e^{-2\pi ix \omega}f(x) dx$

If we translate $f$ by $\delta$ to get $f'(x)=f(x-\delta)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=e^{-2\pi i\delta \omega}\mathcal{F}(f)(\omega)$. In other words, $\mathcal{F}(f')(\omega)$ is a function of $\mathcal{F}(f)(\omega)$, $\delta$, and $\omega$.

If we scale $f$ by $a$ to get $f'(x)=f(x/a)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=\frac{1}{|a|}\mathcal{F}(f)(\omega/a)$. In other words, $\mathcal{F}(f')(\omega)$ is no longer a function of $\mathcal{F}(f)(\omega)$, but instead is a function of $\mathcal{F}(f)(\omega/a)$.

Does there exist a hypothetical new transformation $\mathcal{T}$, like the Fourier transform, but which obeys the following property: If $f'(x)=f(ax-\delta)$, then $\mathcal{T}(f')(\omega)$ is a function only of $\mathcal{T}(f)(\omega)$, $\omega$, $a$, and $\delta$.

Is there a cousin of the Fourier transform which obeys the following property: if I rescale and translate a function, then its transformed form, at each frequency, only depends on the original transformed function at that frequency? In detail:

Given a function $f$, we can take its Fourier transform to get $\mathcal{F}(f)$:

$\mathcal{F}(f)(\omega)=\int e^{-2\pi ix \omega}f(x) dx$

If we translate $f$ by $\delta$ to get $f'(x)=f(x-\delta)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=e^{-2\pi i\delta \omega}\mathcal{F}(f)(\omega)$. In other words, $\mathcal{F}(f')(\omega)$ is a function of $\mathcal{F}(f)(\omega)$, $\delta$, and $\omega$.

If we scale $f$ by $a$ to get $f'(x)=f(x/a)$, the Fourier transform of $f'$ obeys $\mathcal{F}(f')(\omega)=\frac{1}{|a|}\mathcal{F}(f)(\omega/a)$. In other words, $\mathcal{F}(f')(\omega)$ is no longer a function of $\mathcal{F}(f)(\omega)$, but instead is a function of $\mathcal{F}(f)(\omega/a)$.

Does there exist a hypothetical new transformation $\mathcal{T}$, like the Fourier transform, but which obeys the following property: If $f'(x)=f(ax-\delta)$, then $\mathcal{T}(f')(\omega)$ is a function only of $\mathcal{T}(f)(\omega)$, $\omega$, $a$, and $\delta$.