http://dbpedia-live.openlinksw.com:8891/data/Hausdorff%E2%80%93Young_inequality.atom2020-01-21T01:50:17.729042ZOData Service and Descriptor Documenthttp://dbpedia-live.openlinksw.com/resource/Hausdorff–Young_inequality2020-01-21T01:50:17.729042Z2016-10-24 08:26:29ZWilliam HenryWilliam Henry Young626892195Young2238633241282017-12-09 01:17:23Z1913In mathematics, the Hausdorff−Young inequality bounds the Lq-norm of the Fourier coefficients of a periodic function for q ≥ 2. William Henry Young (1913) proved the inequality for some special values of q, and Hausdorff (1923) proved it in general.Hausdorff–Young inequality2016-09-09 03:03:36Z14In mathematics, the Hausdorff−Young inequality bounds the Lq-norm of the Fourier coefficients of a periodic function for q ≥ 2. William Henry Young (1913) proved the inequality for some special values of q, and Hausdorff (1923) proved it in general. More generally the inequality also applies to the Fourier transform of a function on a locally compact group, such as Rn, and in this case Babenko (1961) and Beckner (1975) gave a sharper form of it called the Babenko–Beckner inequality.We consider the Fourier operator, namely let T be the operator that takes a function f {\displaystyle f} on the unit circle and outputs the sequence of its Fourier coefficients f ^ ( n ) = 1 2 π ∫ 0 2 π e − i n x f ( x ) d x , n = 0 , ± 1 , ± 2 , … . {\displaystyle {\widehat {f}}(n)={\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-inx}f(x)\,dx,\quad n=0,\pm 1,\pm 2,\dots .} Parseval's theorem shows that T is bounded from L 2 {\displaystyle L^{2}} to ℓ 2 {\displaystyle \ell ^{2}} with norm 1. On the other hand, clearly, | ( T f ) ( n ) | = | f ^ ( n ) | = | 1 2 π ∫ 0 2 π e − i n t f ( t ) d t | ≤ 1 2 π ∫ 0 2 π | f ( t ) | d t {\displaystyle |(Tf)(n)|=|{\widehat {f}}(n)|=\left|{\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-int}f(t)\,dt\right|\leq {\frac {1}{2\pi }}\int _{0}^{2\pi }|f(t)|\,dt} so T is bounded from L 1 {\displaystyle L^{1}} to ℓ ∞ {\displaystyle \ell ^{\infty }} with norm 1. Therefore we may invoke the Riesz–Thorin theorem to get, for any 1 < p < 2 that T, as an operator from L p {\displaystyle L^{p}} to ℓ q {\displaystyle \ell ^{q}} , is bounded with norm 1, where 1 p + 1 q = 1. {\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1.} In a short formula, this says that ( ∑ n = − ∞ ∞ | f ^ ( n ) | q ) 1 / q ≤ ( 1 2 π ∫ 0 2 π | f ( t ) | p d t ) 1 / p . {\displaystyle \left(\sum _{n=-\infty }^{\infty }|{\widehat {f}}(n)|^{q}\right)^{1/q}\leq \left({\frac {1}{2\pi }}\int _{0}^{2\pi }|f(t)|^{p}\,dt\right)^{1/p}.} This is the well known Hausdorff–Young inequality. For p > 2 the natural extrapolation of this inequality fails, and the fact that a function belongs to L p {\displaystyle L^{p}} , does not give any additional information on the order of growth of its Fourier series beyond the fact that it is in ℓ 2 {\displaystyle \ell ^{2}} .2018-05-01 03:57:55Z41142014-09-24 13:00:17Z2017-09-30 22:21:30Z814474478