TITLE: Uncertainty principle
QUESTION [14 upvotes]: 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?

REPLY [23 votes]: 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.