What does the Fourier transform of an L-infinity function look like locally? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T10:14:51Zhttp://mathoverflow.net/feeds/question/25532http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/25532/what-does-the-fourier-transform-of-an-l-infinity-function-look-like-locallyWhat does the Fourier transform of an L-infinity function look like locally?André Henriques2010-05-21T21:50:15Z2010-05-22T01:00:51Z
<p><b>Question:</b>
What does an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ look like locally?</p>
<p>As formulated, the question might be a bit difficult to answer since the Fourier transform of a function <i>f</i> ∈ <i>L</i><sup>∞</sup>(ℝ) is a distribution, and it is not easy to "write down" a distribution.
So let me first illustrate the situation at hand with an easy example:</p>
<blockquote>
<p><em><b>Example:</b></em>
The Fourier transform of the Heaviside function <i>H</i>(x) (i.e. the characteristic function of the positive reals) is given by
a linear combination of the function 1/x and of the Dirac delta function
(see <a href="http://en.wikipedia.org/wiki/Heaviside_step_function#Fourier_transform" rel="nofollow">this</a> Wikipaedia entry for the exact formula, as well as for the meaning of the distribution "1/x").</p>
</blockquote>
<p>The formalism of distributions is bit overkill for talking about measures, and things that look like 1/x.
For example, the primitive of an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ is
always a function (well defined outside of a set of measure zero).
Using the above observation, we get the following</p>
<p><b>Reformulation of the question:</b><br>
Let <i>f</i>(x) ∈ <i>L</i><sup>∞</sup>(ℝ) be a function, and let <i>g</i>(x) be a primitive of its Fourier transform.<br>
• What can <i>g</i>(x) look like locally?<br>
• What local conditions must <i>g</i> satisfy in order to have a chance of coming from some
<i>f</i> ∈ <i>L</i><sup>∞</sup>(ℝ)?<br>
• On what kind of sets can <i>g</i> fail to be continuous?<br></p>
http://mathoverflow.net/questions/25532/what-does-the-fourier-transform-of-an-l-infinity-function-look-like-locally/25544#25544Answer by fedja for What does the Fourier transform of an L-infinity function look like locally?fedja2010-05-22T01:00:51Z2010-05-22T01:00:51Z<p>It is pretty much the same as to describe the class $G$ of functions $g$ on the circle whose Fourier coefficients decay as $O(|k|^{-1})$. There is no nice "space side" property $P$ that would characterize them but for every nice "space side" property $P$ one can figure out in finite time if it holds for all such functions or not.</p>
<p>As to your particular questions, the answers are</p>
<p>1) On the circle being in this class it is a local property (this needs compactness of the circle) because if $g\in G$, then the product of $g$ and any sufficiently smooth function is in $G$ and you can do partitions of unity.</p>
<p>2) There are obvious inclusions $BV\subset G\subset BMO$ ("bounded variation" and "bounded mean oscillation"). If you need something tighter than that, tell the family of comparison spaces you want to use.</p>
<p>3) Since <code>$\sum_{k\ge 1} \frac 1kz^k$</code> is unbounded at $1$ and continuous everywhere else, we can move such spikes around to create a function that is locally unbounded on any closed set we want and discontinuous on any <code>$F_\sigma$</code> set we want including the entire circle.</p>