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Hilbert transform of a function $g(t)$ which is defined in time domain, would result in another function in time domain. Is there any other transformation like Hilbert that the results be in time domain?

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    $\begingroup$ i wish we could discourage or deprecate the use of the symbol "$f(\cdot)$" for functions and leave the symbol "$f$" for ordinary frequency. that is $$f \triangleq \frac{\Omega}{2 \pi}$$ $\endgroup$ – robert bristow-johnson May 12 at 16:26
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TLDR: if the variable $t$, and the equivalent in the expression of the kernel have the same homogeneity, (I believe that) you will get a time-domain transformation.

Here comes the long version. In the continuous domain with $t\in\mathbb{R}$, a large number of practical or known linear transformations (not called transforms yet) can be expressed as, for "many" functions $g(t)$ and a fixed bivariate kernel $K(t,u)$, as: the integral of their product over $t$:

$$ g_K(u) = \alpha_K\int_{-\infty}^{\infty}g(t) K(t,u)\mathrm{d}t$$

Those specific linear transformations belong to a wider concept of integral operators, that act function spaces. The constant $\alpha_K$ is there for normalization purposes, I will skip it from now on.

Such operators are a central piece of analysis, and they exist under several technical conditions (for instance, those with finite energy are Hilbert-Schmidt operators). For us signal/image processing people, there are some examples, with different notations for the second variable:

  • $K(t,f) = e^{-2j\pi tf}$: Fourier transform
  • $K(t,s) = e^{-s t}$: Laplace transform
  • $K(t,\tau) = \frac{1}{t-\tau}$: Hilbert transform (as special case of potential integral operators, with a Cauchy kernel)
  • $K(t,\tau) = e^{-\frac{(\tau-t)^2}{4}}$: Gauss-Weierstrass transform
  • $K(t,\tau) = h(\tau-t)$: convolution with filter defined by $h$ impulse response

Similar expressions (some in higher dimension) can be derived for wavelet transforms, cosine transform, etc.
I don't know of a precise definition about the difference between "transformations" and "transform". In cases I use, I prefer the notion of transform when the representation preserves most of the original information in the function, and especially when it is inversible, or when there exists a kernel $\kappa$, such that:

$$ g(t) \sim \alpha_\kappa\int_{-\infty}^{\infty}g_K(u) \kappa(t,u)\mathrm{d}u$$

In the Fourier or Laplace transforms, $(t,f)$ or $(t,s)$ couples appear as products in the kernel, so the dual variables can be seen as inverses of each other. They have reciprocal units (like second and Hertz) In the convolution, Gauss and Hilbert transforms cases, $(t,\tau)$ appears in a difference (it could be a sum as well). So they should have the same physical unit. Resultantly, the output function $g_K(u)$ resides in the same time/temporal domain as $g(t)$.

So I would say: as long as $t$ and $\tau $ appear in the same homogeneous manner in the kernel $K(t,\tau)$, you will get a time-domain transformation.

Honestly, I don't know about generic theorems around that.

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    $\begingroup$ i wish we could discourage or deprecate the use of the symbol "$f(\cdot)$" for functions and leave the symbol "$f$" for ordinary frequency in the continuous-time, continuous-frequency Fourier transform. that is $$f \triangleq \frac{\Omega}{2 \pi}$$ $\endgroup$ – robert bristow-johnson May 12 at 16:27
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    $\begingroup$ if you change "$f(t)$" to "$x(t)$" in your answer, i will change the OP notation. but it should be consistent. $\endgroup$ – robert bristow-johnson May 12 at 16:31
  • $\begingroup$ Would by fanctioned for talking about sunction? $\endgroup$ – Laurent Duval May 12 at 16:43
  • $\begingroup$ why did you use "$s$" for goodness sake? $\endgroup$ – robert bristow-johnson May 12 at 16:56
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This question is hard to answer because it depends on your definition of "transform". The Hilbert transform can be written as a convolution with the kernel

$$h(t)=\frac{1}{\pi t}\tag{1}$$

and, consequently, it can be represented by a linear time-invariant (LTI) system with an impulse response given by $(1)$. So the application of any LTI system to a signal can be seen as a transform or mapping which doesn't change the domain of the signal. E.g., the application of an ideal band pass filter to an input signal maps that signal to the space of band-limited functions, and in this sense, it is a transform(ation).

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