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Suppose $\Omega \subset \mathbb{R}^d$ is a bounded, simply-connected domain with whatever other "niceness" properties are necessary for this question to make sense. Define the "Fourier transform" over $\Omega$ of a function $u(x)$, where $x\in\Omega$, as $$ \hat{u}(q) = \int_{\Omega} dx\, \mathrm{e}^{-i q\cdot x} u(x).\qquad (1) $$ Can we define a sensible inverse of this transform?

Clearly, if $\Omega = \mathbb{R}^d$ then we have the inverse $$ u(x) = \frac{1}{{(2\pi)}^d}\int_{\mathbb{R}^d} dq\, \mathrm{e}^{+i q\cdot x} \hat{u}(q). $$ By analogy with this, we might expect an inverse transform of the form $$ u(x) = \int_{\mathbb{R}^d} dq\, K(q, x)\, \hat{u}(q)\qquad (2) $$ for some kernel $K$ to be determined. Substituting (2) into (1), one obtains the following condition on $K$: $$ \int_{\Omega} dx\, \mathrm{e}^{-i q\cdot x}\, K(q', x) \;=\; \delta(q - q'). $$ I'm not quite sure where to go from here. Am I on the right track with this sort of thinking?

Some reading has led me to suspect that $K$, if it exists, will turn out to be some kind of infinite sum of products of eigenfunctions defined on $\Omega$.

Thanks for any and all help.

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2 Answers 2

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Denote by $\mathcal{F}$ the Fourier transform in $\mathbb{R}^n$. Given $u\in L^1(\Omega)$ define its $\Omega$-Fourier transform as $$ \mathcal{F}_{\Omega}u(q)=\int_\Omega u(x)\,e^{-iqx}\,dx. $$ Let $\tilde u\colon\mathbb{R}^n\to\mathbb{R}$ be $$ \tilde u(x)=\begin{cases}u(x) & \text{if }x\in\Omega,\\0 & \text{if }x\not\in\Omega.\end{cases} $$ Then $$ \mathcal{F}_{\Omega}=\mathcal{F}\tilde u. $$ It follows that $$ \mathcal{F}_{\Omega}^{-1}(\mathcal{F}_{\Omega}u)=\mathcal{F}^{-1}(\mathcal{F}_{\Omega}u)\Bigr|_\Omega. $$

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$$ \int_\Omega f =\int_{\Re^n} f \chi_\Omega$$

where $\chi_\Omega$ is the charcteristic function of the domain $\Omega$. Then the inverse can be given as a convolution of the function with the characteristic function.

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  • $\begingroup$ Well, I considered that, but I wasn't able to extract an inverse of the given integral transform that didn't run around in circles. Can you be more specific? $\endgroup$ Jan 28, 2015 at 22:08

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