Range of the Fourier transform on L^1 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:30:17Z http://mathoverflow.net/feeds/question/8085 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1 Range of the Fourier transform on L^1 user17240 2009-12-07T07:22:49Z 2010-02-01T06:36:23Z <p>It is well known that the Fourier transform $\mathcal{F}$ maps $L^1(\mathbb{R}^d)$ into, but not onto, $\overline{C_0^0}(\mathbb{R}^d)$, where the closure is taken in the $L^\infty$ norm. This is a consequence of the open mapping theorem, for instance. </p> <p>My question is: what's an explicit example of a function in $\overline{C_0^0}(\mathbb{R}^d)$ which is not in the image of $L^1(\mathbb{R}^d)$ under the Fourier transform? </p> <p>I would also like to know whether there is a useful characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$. </p> <p>Remark: it is easy to see that the Banach space $\overline{C_0^0}(\mathbb{R}^d)$ consists of all continuous functions $f$ on $\mathbb{R}^d$ such that $f(\xi)\rightarrow 0$ as $|\xi|\rightarrow\infty$.</p> <p>Thank you!</p> http://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1/8102#8102 Answer by Gian Maria Dall'Ara for Range of the Fourier transform on L^1 Gian Maria Dall'Ara 2009-12-07T13:15:16Z 2010-02-01T06:36:23Z <p>For the first question I commented above that the function $\sqrt(1-|x|^2)$ extended to $0$ outside the unit ball is not the fourier transform of any integrable function in dimension 2 or higher. In dimension $1$ there's "Further results" of Chapter I in Introduction to Fourier Analysis of Stein. In case you don't have access to the book, this is the construction: Observe that $|\int_a^b \sin(x)/x\ dx| \leq C&lt;\infty$ for any strictly positive $a$ and $b$. Now, if $f\in L^1$ and $F(f)$ is odd you have $F(f)(x) = \int f(t) \sin(xt)\ dt$ up to a multiplicative constant. Than it's easy to see from the previous estimate that $|\int_1^b F(f)(x)/xdx|\leq C'&lt;\infty$ uniformly in $b$. So a function which is continuous, odd and which decays too slowly ($1/\log(x)$ will do) is not the Fourier transform of an integrable function.</p> http://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1/8397#8397 Answer by Yemon Choi for Range of the Fourier transform on L^1 Yemon Choi 2009-12-09T23:47:01Z 2009-12-09T23:52:33Z <p>It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from $L^1$(sensible measure space) onto $C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can <em>know</em> the answer to "is the FT onto?" <em>must</em> be "no", before looking for an example or using properties of the Fourier transform.</p> <p>(My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.)</p> <p>Anyway. I claim that there is no continuous linear bijection between $L^1({\mathbb R}^d)$ and $C_0(X)$, where $X$ is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that</p> <p>(i) every bounded linear operator from $C_0(X)$ to $L^1({\mathbb R}^d)$ is <em>weakly compact</em>;</p> <p>(ii) if the identity map on a Banach space $E$ is weakly compact, then $E$ is <a href="http://en.wikipedia.org/wiki/Reflexive%5Fspace" rel="nofollow">reflexive</a>;</p> <p>(iii) $L^1({\mathbb R}^d)$ is not reflexive (ibid).</p> <p>Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather <a href="http://en.wikipedia.org/wiki/Grothendieck%5Fconstant" rel="nofollow">powerful, fundamental and beautiful result</a> - due to some promising former student of Dieudonn&eacute; and Schwartz, not sure if he ever went on to do anything important...)</p>