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I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

$$\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx$$

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: $$h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))$$

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: $$h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))$$ What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

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    $\begingroup$ It seems rather ambitious to expect an inverse for a map from functions in one variable to functions in 2 variables. $\endgroup$
    – S. Carnahan ♦
    Commented Jul 26, 2011 at 8:31
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    $\begingroup$ It doesn't look that ambitious to look for an inverse if you have an injective map and want to compute the antecedent of an element which is known to be in the image! $\endgroup$
    – Julien Puydt
    Commented Jul 26, 2011 at 16:06
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    $\begingroup$ I guess $a$ and $b$ are real? Have you already exploited a series expansion $e^{\pm 2(ax\pm b)}$ for $\tanh$. Then this becomes essentially a question, which is equivalent to recover a probability distribution in terms of moment sums, from something like $2 +2 \sum_{n>0}EX^n e^{bn}$, which is holomorph in $b$ and where $EX^n$ is the $n$ th moment of your distribution. So in principle you can recover the moments $EX^n$, which is sufficient for the distribution of $X$, if it is distributed with total mass $1$. So $(3)$ yes! $\endgroup$
    – Marc Palm
    Commented Jul 26, 2011 at 17:05
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    $\begingroup$ Indeed, $\overline{f}(a,b)$ is essentially the Radon transform of the two-variable function $f(x) \tanh(y)$, so the inverse Radon transform will do the trick. If one is only given $\overline{f}(1,b)$, this is essentially a convolution of f with the distribution $\tanh$, so a Fourier transform should be able to perform deconvolution as long as one avoids frequencies where the Fourier transform of $\tanh$ (or its derivative $\sech^2$, which is integrable) vanishes. $\endgroup$
    – Terry Tao
    Commented Jul 26, 2011 at 18:25
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    $\begingroup$ Sorry, I did not directly see the question in your comment and my previous comment was also partly wrong. You are not getting moments, but exponential moments. First, $tanh(z) = 1-2 (1+ e^{2z})^{-1} = 1 + 2\sum\limits_{n>0} (-e^{2z})^n$ for $z<0$ and similarly for $x>0$. Then plugin $z=b+x$ and integrate summand for summand. So you get some Fourier expansion $\sum\limits_{n>0} a_n e^{bn}$ for $b<0$. Since $a_n$ is essentially the Laplace transform of $f$ at $n$, and I know at least that the question to recover $f$ from exponential moments should be discussed in the standard books. $\endgroup$
    – Marc Palm
    Commented Jul 26, 2011 at 19:51

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