Do you know this form of an uncertainty principle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:03:54Z http://mathoverflow.net/feeds/question/75925 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75925/do-you-know-this-form-of-an-uncertainty-principle Do you know this form of an uncertainty principle? Dirk 2011-09-20T08:50:54Z 2011-11-20T22:20:44Z <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#75930 Answer by Piero D'Ancona for Do you know this form of an uncertainty principle? Piero D'Ancona 2011-09-20T09:47:19Z 2011-09-20T09:47:19Z <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#81462 Answer by Antoine Levitt for Do you know this form of an uncertainty principle? Antoine Levitt 2011-11-20T22:20:44Z 2011-11-20T22:20:44Z <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>