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.

You probably know the Heisenberg uncertainty principle: For any function $g\in L^2(\mathbb{R})$ for which the respective expressions exist it holds that $$ \frac{1}{4}\|g\|_2^4 \leq \int_{\mathbb{R}} |x|^2 |g(x)|^2 dx \int_{\mathbb{R}} |g'(x)|^2 dx. $$

This inequality is not only important in quantum mechanics, but also in signal processing for the short-time Fourier transform, see here.

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.

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 $$ \|g\|_2\leq \|xg'\|_2. $$

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 two blog posts). Interestingly, the Heisenberg uncertainty principle derives from the short time Fourier transform and the last "uncertainty principle" derives from the wavelet transform.

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.

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?

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    $\begingroup$ Isn't the last inequality (I assume you meant $\|g\|_2 \leq \|xg'\|_2$) just a manifestation of Hardy's inequality? en.wikipedia.org/wiki/Hardy%27s_inequality In which case, yes, people are familiar with it. $\endgroup$ Sep 20, 2011 at 13:37
  • $\begingroup$ In 1D Hardy's inequality is a bit awkward to express and you should work with functions vanishing at zero to get a neat formulation. Of course you can go from one to the other and back, but I would not say it is an instant consequence $\endgroup$ Sep 20, 2011 at 15:53

2 Answers 2


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).

  • $\begingroup$ Wow, there are lot of constants involved. As far as I've seen, it seems that the case $\gamma=0$, $a=1$ and $\alpha=-1$ is not included? $\endgroup$
    – Dirk
    Sep 20, 2011 at 10:24
  • $\begingroup$ You can take $n=1,a=1/2,\beta=\gamma=0,\alpha=-1$ to get your inequality, if I'm not mistaken $\endgroup$ Sep 20, 2011 at 10:34
  • $\begingroup$ I got a little bit confused with all these different inequalities around (many of them called Caffarelli-Kohn-Nirenberg) but now I see. That's an interesting relation though. The Heisenberg uncertainty is also included? ($\gamma=\alpha=0$, $\beta=−1$, $a=1/2$) $\endgroup$
    – Dirk
    Sep 20, 2011 at 12:01
  • $\begingroup$ Yes indeed, and it's not limited to the $L^2$ framework as you see $\endgroup$ Sep 20, 2011 at 15:47

I find the neatest "standard" uncertainty principle is the one with commutators, see e.g. http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm. I think that readily gives both your inequalities.

  • $\begingroup$ Of course the first one is an instance of the abstract "Robinson's uncertainty principle". However, for the second it's not that straightforward... $\endgroup$
    – Dirk
    Nov 21, 2011 at 6:50

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