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I need to know if one can view a function $f\in L^{\infty}(\mathbb{R})$ as a Fourier transform of a certain function, say g?

If the answer is positive please state the proof, or help me find one. Thanks

Jean-Claude Arbaut
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user157524
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1 Answers1

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In general, no. For example, take $f(\xi) = 1$. There is a generalization of the Fourier transform, the Fourier–Stieltjes transform, and that maps a Dirac delta to $f(\xi) = 1$. So, in general, this inverse could be a distribution.

$L^1(\mathbb{R})$ is pretty much the largest class of functions which we can define the Fourier transform for. The Fourier transform of any function in $L^1(\mathbb{R})$ must go to zero at $\pm \infty$.

Edit to make that last point clearer: any function which does not go to zero at infinity will not be the Fourier transform of some function.

Another edit: See the comments; saying "$L^1(\mathbb{R})$ is the largest class..." was careless.

Jean-Claude Arbaut
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Tim Carson
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    There is a well-defined way to extend the Fourier transform to L^2 functions, using a limiting argument, such that the definition is consistent with the integral for functions which are also in L^1. Indeed, the L^2 Fourier transform is an isometry (and therefore a bijection) from L^2->L^2. –  Aug 17 '14 at 04:14
  • Ah yes, I was wrong when I said "largest", since $L^2(\mathbb{R})$ and $L^1(\mathbb{R})$ have no inclusion between them. – Tim Carson Aug 17 '14 at 04:21
  • But, it somehow feels like the closest to being able to reach all of $L^\infty(\mathbb{R})$. – Tim Carson Aug 17 '14 at 04:23
  • The Fourier transform of an $L^1$ function is continuous, whereas the Fourier transform of an $L^2$ function need not be. Example: sin(x)/x is transformed to a rectangular "pulse". So we can reach more $L^\infty$ functions by extending the transform to $L^1 \cup L^2$, although I'm not sure how to quantify how many more. We'll still miss some, such as any nonzero constant function. If we extend the transform to the space of tempered distributions, then it is once again a bijection on that space. But not every $L^\infty$ function can be identified with a distribution, so even that is not enough. –  Aug 17 '14 at 04:37