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Let be $q>p$ then $L^q(\Omega)\subset L^p(\Omega)$. I will be able to say that all $f \in L^q$, such that $q>1$, have a Fourier Transform?.

pdta:I asking this because I am read that exist Fourier transform only when $f \in L^1$ but $L^q \subset L^p$.

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On which space are you working? With which measure? – Davide Giraudo Oct 30 '12 at 16:07
In the Lebesgue space – user46060 Oct 30 '12 at 16:22
And what is $\Omega$? – Davide Giraudo Oct 30 '12 at 16:28
The inclusion between $L^p$ spaces is only valid for $\Omega$ with finite measure (in this case: the larger the exponent, the smaller the space). I think this is what @Davide Giraudo is after... – Dirk Oct 30 '12 at 16:36
$\Omega$ is bounded – user46060 Oct 30 '12 at 16:39

Given a Banach space with a multiplication (usually convolution in $L^p$-spaces) such that $$\|x\cdot y\|\leq C \|x\|\cdot \|y\|$$ we do have a Gelfand transform which on $L^1(\mathbb{R})$ coincides with the Fourier transform or in the case of $L^1(\mathbb{T})$ the Fourier coefficients (here $\mathbb{T}=\mathbb{R}/\mathbb{Z}\sim(-\pi,\pi]$ is the circle group).

That $\Omega$ is bounded is not really sufficient in order to have a worthy Fourier transform. Surely, for bounded $\Omega$ and $f\in L^p(\Omega)$ we may define $$\hat{f}(\xi)=\int_\Omega f(x)e^{-ix\xi}dx$$ Now, by defining $f=0$ on $\mathbb{R}\setminus\Omega$ we extend $f$ to $L^p(\mathbb{R})$, and then we may consider a translate of $f$, i.e. $f_a(x)=f(x-a)$, for which we have $$\int_\mathbb{R} f_a(x)e^{-ix\xi}dx=\int_\mathbb{R} f(x-a)e^{-ix\xi}dx=\\\int_\mathbb{R} f(x)e^{-i(x+a)\xi}dx=e^{-ia\xi}\int_\mathbb{R} f(x)e^{-ix\xi}dx=e^{-ia\xi}\hat{f}(\xi)$$ This is a problem because if $f$ is nice enough, we would have an inverse Fourier transform and the translate must then belong to $L^p(\Omega)$ - but that is not the case unless we have some additional algebraic structure on $\Omega$.

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First, the Fourier transform exists not only when $f \in L^1$; the Fourier transform can also be defined in the space of $L^2$ and some other spaces, like spaces of generalized functions corresponding to tempered distributions.

Second, $L^q(\Omega)$ is not necessary a subspace of $L^p(\Omega)$ if $\Omega$ is unbounded.

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