Editing Entropic uncertainty (section)

==Sketch of proof==
The proof of this tight inequality depends on the so-called '''(''q'',&nbsp;''p'')-norm''' of the Fourier transformation.  (Establishing this norm is the most difficult part of the proof.)

From this norm, one is able to establish a lower bound on the sum of the (differential) [[Rényi entropy|Rényi entropies]], {{math| ''H<sub>α</sub>({{!}}f{{!}}²)+H<sub>β</sub>({{!}}g{{!}}²)'' }}, where {{math|''1/α + 1/β'' {{=}} 2}}, which generalize the Shannon entropies.  For simplicity, we consider this inequality only in one dimension; the extension to multiple dimensions is straightforward and can be found in the literature cited.

===Babenko–Beckner inequality===
The '''(''q'',&nbsp;''p'')-norm''' of the Fourier transform is defined to be<ref name=Bialynicki>{{Cite journal | doi = 10.1103/PhysRevA.74.052101| title = Formulation of the uncertainty relations in terms of the Rényi entropies| journal = Physical Review A| volume = 74| issue = 5| page = 052101| year = 2006| last1 = Bialynicki-Birula | first1 = I. |arxiv = quant-ph/0608116 |bibcode = 2006PhRvA..74e2101B | s2cid = 19123961}}</ref>

:<math>\|\mathcal F\|_{q,p} = \sup_{f\in L^p(\mathbb R)} \frac{\|\mathcal Ff\|_q}{\|f\|_p},</math>   where <math>1 < p \le 2~,</math> &nbsp; and <math>\frac 1 p + \frac 1 q = 1.</math>

In 1961, Babenko<ref>K.I. Babenko.  ''An inequality in the theory of Fourier integrals.'' Izv. Akad. Nauk SSSR, Ser. Mat. '''25''' (1961) pp. 531&ndash;542 English transl., Amer. Math. Soc. Transl. (2) '''44''', pp. 115-128</ref> found this norm for ''even'' integer values of ''q''.  Finally, in 1975,
using [[Hermite functions]] as eigenfunctions of the Fourier transform, Beckner<ref name=Beckner/> proved that the value of this norm (in one dimension) for all ''q''  ≥  2  is
:<math>\|\mathcal F\|_{q,p} = \sqrt{p^{1/p}/q^{1/q}}.</math>
Thus we have the '''[[Babenko–Beckner inequality]]''' that
:<math>\|\mathcal Ff\|_q \le \left(p^{1/p}/q^{1/q}\right)^{1/2} \|f\|_p.</math>

===Rényi entropy bound===
From this inequality, an expression of the uncertainty principle in terms of the [[Rényi entropy]] can be derived.<ref name=Bialynicki/><ref>H.P. Heinig and M. Smith, ''Extensions of the Heisenberg–Weil inequality.'' Internat. J. Math. & Math. Sci., Vol. 9, No. 1 (1986) pp. 185&ndash;192. [http://www.hindawi.com/GetArticle.aspx?doi=10.1155/S0161171286000212]</ref>

Let <math>g=\mathcal Ff, \, 2\alpha=p, \, 2\beta=q,</math> so that <math>\frac1\alpha+\frac1\beta=2</math> and <math>\frac12\le\alpha\le1\le\beta</math>, we have
:<math>\left(\int_{\mathbb R} |g(y)|^{2\beta}\,dy\right)^{1/2\beta}
       \le \frac{(2\alpha)^{1/4\alpha}}{(2\beta)^{1/4\beta}}
       \left(\int_{\mathbb R} |f(x)|^{2\alpha}\,dx\right)^{1/2\alpha}.
</math>
Squaring both sides and taking the logarithm, we get
:<math>\frac 1\beta \log\left(\int_{\mathbb R} |g(y)|^{2\beta}\,dy\right)
       \le \frac 1 2 \log\frac{(2\alpha)^{1/\alpha}}{(2\beta)^{1/\beta}}
       + \frac 1\alpha \log \left(\int_{\mathbb R} |f(x)|^{2\alpha}\,dx\right).
</math>

We can rewrite the condition on <math>\alpha, \beta</math> as

:<math>\alpha(1-\beta)+\beta(1-\alpha)=0</math> 

Assume <math>\alpha,\beta\ne1</math>, then we multiply both sides by the negative 
:<math>\frac{\beta}{1-\beta}=-\frac{\alpha}{1-\alpha}</math> 
to get
:<math>\frac {1}{1-\beta} \log\left(\int_{\mathbb R} |g(y)|^{2\beta}\,dy\right)
       \ge \frac\alpha{2(\alpha-1)}\log\frac{(2\alpha)^{1/\alpha}}{(2\beta)^{1/\beta}}
       - \frac{1}{1-\alpha} \log \left(\int_{\mathbb R} |f(x)|^{2\alpha}\,dx\right) ~.
</math>

Rearranging terms yields  an inequality in terms of the sum of Rényi entropies,
:<math>\frac{1}{1-\alpha} \log \left(\int_{\mathbb R} |f(x)|^{2\alpha}\,dx\right)
       + \frac {1}{1-\beta} \log\left(\int_{\mathbb R} |g(y)|^{2\beta}\,dy\right)
       \ge \frac\alpha{2(\alpha-1)}\log\frac{(2\alpha)^{1/\alpha}}{(2\beta)^{1/\beta}};
</math>
:<math> H_\alpha(|f|^2) + H_\beta(|g|^2) \ge \frac 1 2 \left(\frac{\log\alpha}{\alpha-1}+\frac{\log\beta}{\beta-1}\right) - \log 2</math>

====Right-hand side====

:<math>\frac\alpha{2(\alpha-1)}\log\frac{(2\alpha)^{1/\alpha}}{(2\beta)^{1/\beta}}</math>

:<math>=\frac12\left[\frac{\alpha}{\alpha-1}\log(2\alpha)^{1/\alpha} + \frac{\beta}{\beta-1}\log(2\beta)^{1/\beta}\right]</math>

:<math>=\frac12\left[\frac{\log2\alpha}{\alpha-1} + \frac{\log2\beta}{\beta-1}\right]</math>

:<math>=\frac12\left[\frac{\log\alpha}{\alpha-1} + \frac{\log\beta}{\beta-1}\right] + \frac12\log2\left[\frac{1}{\alpha-1} + \frac{1}{\beta-1}\right]</math>

:<math>=\frac12\left[\frac{\log\alpha}{\alpha-1} + \frac{\log\beta}{\beta-1}\right] + \frac12\log2\left[\frac{1}{\alpha-1} + \frac{1}{\beta-1} - \frac{\alpha}{\alpha-1} - \frac{\beta}{\beta-1}\right]</math>

:<math>=\frac12\left[\frac{\log\alpha}{\alpha-1} + \frac{\log\beta}{\beta-1}\right] + \frac12\log2\left[-2\right]</math>

:<math>=\frac12\left[\frac{\log\alpha}{\alpha-1} + \frac{\log\beta}{\beta-1}\right] - \log2</math>

===Shannon entropy bound===
Taking the limit of this last inequality as <math>\alpha, \, \beta \to 1</math> and the substitutions <math>\Alpha=\alpha-1, \Beta=\beta-1</math> yields the less general Shannon entropy inequality,
:<math>H(|f|^2) + H(|g|^2) \ge \log\frac e 2,\quad\textrm{where}\quad g(y) \approx \int_{\mathbb R} e^{-2\pi ixy}f(x)\,dx~,</math>
valid for any base of logarithm, as long as we choose an appropriate unit of information, [[bit]], [[Nat (unit)|nat]], etc.

The constant will be different, though, for a different normalization of the Fourier transform, (such as is usually used in physics, with normalizations chosen so that ''ħ''=1 ), i.e.,
:<math>H(|f|^2) + H(|g|^2) \ge \log(\pi e)\quad\textrm{for}\quad g(y) \approx \frac 1{\sqrt{2\pi}}\int_{\mathbb R} e^{-ixy}f(x)\,dx~.</math>
In this case, the dilation of the Fourier transform absolute squared by a factor of 2{{mvar|π}}  simply adds log(2{{mvar|π}}) to its entropy.