# Uncertainty principle

A version of the uncertainty principle says that a function and its Fourier transform cannot be both with compact support: it is not difficult to prove since a compactly supported distribution has an entire Fourier transform. Another version is that $$\left\Vert{\frac{du}{dx}}\right\Vert_{L^2(\mathbb R)} \left\Vert{xu}\right\Vert_{L^2(\mathbb R)}\ge \frac12\left\Vert{u}\right\Vert_{L^2(\mathbb R)}^2,$$ and many other quantitative versions are available for functions bounded above by Gaussians functions as well as their Fourier transform.

All this seems to be compatible with the existence of a function $u$ in $L^2(\mathbb R)$ such that $$\text{support u}\subset \mathbb R_+,\quad \text{support \hat u}\subset \mathbb R_+.$$ My question: is there an "explicit" example of such a function?

• You know, if you create three brand new tags for your question, it means that nobody is actively following these tags, so nobody is going to see this question highlighted in their favorite tags. Not to mention, all three tags are terrible. I don't know enough to tell you anything about the question, or about the correct tags to use, but you might want to look at the Tags page and find a better tag for this question, so it will get better exposure. Oct 19, 2016 at 13:34
• I took the liberty and modified the tags - I guess these are appropriate, but feel free to change…
– Dirk
Oct 19, 2016 at 14:03

Yes, there is one such example: $u \equiv 0$.
The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).
By Titchmarsh's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.
If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.