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
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ Is the constant function $1$ a Fourier transform of a function? What kind of function is $g$ supposed to be? $\endgroup$– Jonas MeyerAug 17, 2014 at 4:05
-
1$\begingroup$ The Fourier transform of any L^1 function is uniformly continuous and its tails decay to zero, so there are many counterexamples available. $\endgroup$– user169852Aug 17, 2014 at 4:11
-
$\begingroup$ This may be useful $\endgroup$– InquisitiveAug 17, 2014 at 5:54
-
$\begingroup$ @user157524: see my answer and do not worry about the downvote. It answers your question. Think carefully. $\endgroup$– Mhenni BenghorbalAug 19, 2014 at 19:03
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
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.
-
1$\begingroup$ 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. $\endgroup$– user169852Aug 17, 2014 at 4:14
-
$\begingroup$ Ah yes, I was wrong when I said "largest", since $L^2(\mathbb{R})$ and $L^1(\mathbb{R})$ have no inclusion between them. $\endgroup$ Aug 17, 2014 at 4:21
-
$\begingroup$ But, it somehow feels like the closest to being able to reach all of $L^\infty(\mathbb{R})$. $\endgroup$ Aug 17, 2014 at 4:23
-
$\begingroup$ 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. $\endgroup$– user169852Aug 17, 2014 at 4:37