Do you know this form of an uncertainty principle? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T13:03:54Zhttp://mathoverflow.net/feeds/question/75925http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/75925/do-you-know-this-form-of-an-uncertainty-principleDo you know this form of an uncertainty principle?Dirk2011-09-20T08:50:54Z2011-11-20T22:20:44Z
<p>I hope this question is focused enough - it's not about real problem I have, but to find out if anyone knows about a similar thing.</p>
<p>You probably know the <a href="http://en.wikipedia.org/wiki/Uncertainty_principle" rel="nofollow">Heisenberg uncertainty principle</a>: For any function $g\in L^2(\mathbb{R})$ for which the respective expressions exist it holds that
<code>$$
\frac{1}{4}\|g\|_2^4 \leq \int_{\mathbb{R}} |x|^2 |g(x)|^2 dx \int_{\mathbb{R}} |g'(x)|^2 dx.
$$</code></p>
<p>This inequality is not only important in quantum mechanics, but also in signal processing for the short-time Fourier transform, see <a href="http://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle" rel="nofollow">here</a>.</p>
<p>One can derive this by formally using partial integration
$$
\int_{\mathbb{R}} 1\,|g(x)|^2 dx = -\int_{\mathbb{R}} x\tfrac{d}{dx}|g(x)|^2dx \leq 2\int_{\mathbb{R}} |xg(x)|\,|g'(x)|dx
$$
and Cauchy-Schwarz.</p>
<p>Now, changing just the order of the functions, you obtain this inequality
$$
\int_{\mathbb{R}} |g(x)|^2 dx \leq 2\int_{\mathbb{R}} |xg'(x)|\,|g(x)|dx
\leq \left(\int_{\mathbb{R}} |xg'(x)|^2dx\right)^{1/2}\left(\int_{\mathbb{R}} |g(x)|^2dx\right)^{1/2}
$$
which gives
$$
\|g\|_2\leq \|xg'\|_2.
$$</p>
<p>Ok, this was just playing around. However, this inequality can also be motivated by an abstract consideration about uncertainty principle associated to group-related integral transforms (see my <a href="http://regularize.wordpress.com/2011/09/16/the-uncertainty-principle-for-the-windowed-fourier-transform/" rel="nofollow">two</a> <a href="http://regularize.wordpress.com/2011/09/20/the-uncertainty-principle-for-the-one-dimensional-wavelet-transform/" rel="nofollow">blog
posts</a>). Interestingly, the Heisenberg uncertainty principle derives from the short time Fourier transform and the last "uncertainty principle" derives from the wavelet transform.</p>
<p>The last fact bothers me: In contrast to the fact that both inequalities can be derived from two conceptually very different integral transforms (indeed both underlying groups are very different), they have a very similar formal derivation.</p>
<p>I have the following questions: Is anyone familiar with the last inequality? Could it be useful in any context? Is there some reason why these inequalities seem so entangled?</p>
http://mathoverflow.net/questions/75925/do-you-know-this-form-of-an-uncertainty-principle/75930#75930Answer by Piero D'Ancona for Do you know this form of an uncertainty principle?Piero D'Ancona2011-09-20T09:47:19Z2011-09-20T09:47:19Z<p>There exists a plethora of inequalities relating weighted $L^p$ norms of a function and its derivatives. For instance you have the Caffarelli-Kohn-Nirenberg family of inequalities
$$\| |x|^{-\gamma}u\|_ {L^{r}}\le C \||x|^{-\alpha}\nabla u\|^{a}_ {L^{p}}\||x|^{-\beta}u\|^{1-a}_ {L^{q}}
$$
which hold for a quite large range of parameters; note that here $\alpha,\beta,\gamma$ may assume negative values. You will not have difficulty in googling the vast literature on the subject (let me add that there is a recent paper of mine on arXiv with some improvements on this).</p>
http://mathoverflow.net/questions/75925/do-you-know-this-form-of-an-uncertainty-principle/81462#81462Answer by Antoine Levitt for Do you know this form of an uncertainty principle?Antoine Levitt2011-11-20T22:20:44Z2011-11-20T22:20:44Z<p>I find the neatest "standard" uncertainty principle is the one with commutators, see e.g. <a href="http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm" rel="nofollow">http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm</a>. I think that readily gives both your inequalities.</p>