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I have a function $A : \mathbb{R} \to \mathbb{C}$.

I know there exists unknown function $u: \mathbb{R} \to \mathbb{C}$, such that $A$ is convolution of $u$ and its complex conjugate $A = u * u^*$.

I would like to compute function $|u|^2 = uu^*$. Is there unique solution to this problem and is there efficient computational method how to find it?

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2 Answers 2

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$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.


E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) \\ =\frac54\,u_1(x)-\frac{3ix}{4\sqrt{2\pi}}\,e^{-x^2/2} $$ for real $x$, where $\text{erfi}(x):=\sqrt{\frac2\pi}\int_0^{x\sqrt2}e^{z^2/2}\,dz$.

Here are the graphs of $u_1$ (blue) and $\Re u_2$ (gold) over the interval $[-6,6]$:

enter image description here

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  • $\begingroup$ Thank you for the answer. Apparently I have made a mistake so that I know $B=u(x) * u(-x)$. It is apparent it won't be unique anyway due to possibly different phases in $\hat{u}(t)\hat{u}(t)^*$. Do I fix this ambiguity assuming $u = e^{i \phi(x)}$ and is there closed formula to recover such real function $\phi(x)$? I assume this problem shall be unique but do not know any solution to it. $\endgroup$
    – VojtaK
    Feb 7, 2023 at 16:42
  • $\begingroup$ @VojtaK : You comment is not clear to me. I think such additional questions should be asked in separate, carefully prepared posts. $\endgroup$ Feb 7, 2023 at 16:52
  • $\begingroup$ I have asked related question here math.stackexchange.com/questions/4634301/… $\endgroup$
    – VojtaK
    Feb 7, 2023 at 17:23
  • $\begingroup$ @VojtaK : I am not really active on math.stackexchange.com. $\endgroup$ Feb 7, 2023 at 17:25
  • $\begingroup$ OK, I have migrated the question here mathoverflow.net/questions/440381/… Originally, I thought it will be simpler for this site. $\endgroup$
    – VojtaK
    Feb 7, 2023 at 17:30
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Another counterexample with simpler functions.

I work with even and real functions so their Fourier transform are also even and real functions. Hence, one can omit complex conjugations and replacements of the argument of functions by its opposite.

The functions $u_\pm$ defined by $$u_\pm(x) = (e^{i2x}+e^{-i2x}\pm 1)\frac{\sin^2 x}{x^2} = (2\cos(2x)\pm 1)\frac{\sin^2 x}{x^2}$$ do not have the same modulus but verify $u_+*u_+ = u_-*u_-$.

Indeed, call $f$ the tent function, defined by $f(t) = \max(0,1-|t|)$. Then $f$ is continuous with support $[-1,1]$. Thus the functions $x \mapsto f(x-2) + f(x+2) \pm f(x)$ have the same square and their Fourier transform, which are $u_\pm$. Hence $u_+*u_+ = u_-*u_-$.

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