4   Introduction to Fourier Restriction  Theorem   Hausdorff Young inequality   Assuming that   1   p   2  1 p   1 q   1  Then   Proof  The inequality is true for p   1  q     and for p   2  q   2 the inequality is true since we have equality  Plancherel    For values in between we can interpolate     Are there any other   p   q   for which   We saw that 1   p   2 was necessary  translations   modulations  Khinchin s inequality     Scaling   Plug in f     x     f     x   which is L   normalised    f           f       Then f                 n f         1     which is L 1 normalised    f       1     f     1    So we need for all     0   So we need   n   n q     n p  i e  1 p   1 q   1   Classical questions  What is C p   q the smallest constant such that   f     q   C p   q   f   p   Which functions f satisfy   f     q   f   p   C p   q   2014    Fourier restriction asks which   p   q   permit estimates   f     L q   R       f   L p     n    R is the restricted frequency set  R     n    Example  R   B 1   0       n  unit ball    Basically  sitll governed by Hausdorff Young inequality    Example  R   B 1   0       x n   0    measure 0 subset of   n    R R   p   q   means  Always  R R   1       true for all R   In the second example  only this trivial statement is true  i e  R R   p   q   is false for all other values for p   q    Let R   S n   1 be the unit sphere in   n  Consider   where L q   S n   1   uses the usual surface measure d   on S n   1   Notation  We may call   the  inverse Fourier transform     Let     C c       n      valued    1 2 on B 1   0    with support in B 2   0     May also assume      is   valued         1 on B c   0         bounded           x       m     x   2   1     m   m           x     behaves like   B 1   0   in L p   Consider dilates  f R   x           R   1 x   e 2   i x   v r  R   1   Frequency side   f R             R n   B R   1   0     x    L 1 norm    B R   1   n     S n   1 cap of radius R   1  Spacial side   f R   x         B R   0     x    in L   norm    height 1      n   f R   x     p d x   R n   R n   n   1 q   R n p  n   n   1 q   n p   Consider   g R   x     e i x   w r       R   1 x 1       R   1 x n   1   R   2 x n     Frequency    g r               c y l i n d e r     1   c y l i n d e r          c y l i n d e r     1       R   1   R   1 R   2     1   R n   1   Spatial side    g R     x         c y l i n d e r   x    L   norm        g R   x     p   R n   1   2   R n   1     R n   1   n   1 q   R n   1 q    n   1   n   1 q   n   1 p  Implies B   On Monday  we will build examples h such that h   sees all  of S n   1   R   S n   1   R n   Consider the statement    recall that we called this R S n   1   p   q      Fix     C c       n          B 1   0    For computing L p norms              B c   0     Wave packet function with localised spatial and frequency behaviour   Last time   f R   x     e i x   v R       R   1 x 1       R   1 x n   1   R   2 x n     Frequency    f R              S n   1   f R     q d    Spatial    f R   x          1   f R   p  Note  sphere near n looks like     1   1   1 2         2    S n   1   supp f R     cap of radius R   1   Naive attempt    g R   x     R n       R x    Frequency    g R              S n   1   g R           q d     1  Spatial    g R   x          g R   p   R   n   R n p   Deduce  1   R   n p   n  so   trivial      g R       1 on S n   1 made   g R     L q   S n   1   easy to compute   Could we improve things   Could think about R   1  then   g R     L q   S n   1     1    g R   p   1  This is more efficient  but we can t take a limit  So not so useful   Build a function H   x   which satisfies   H             1 on S n   1   Let       be a maximal collection of   R   1 spaced points on S n   1            R n   1    For each    let A     1     n     n be an affine map which sends   Define           A     H   x                 x     H                           1 on S n   1  actually on R   2 neighbourhood of S n   1      S n   1   H           q d     1   H   x                 x             x      Frequency              x     Spatial   ess supp H       ess supp          bush of tubes     R n   1 many R       R   R 2 tubes in R   1 separated directions   Compute  munder         n         T     x   2   p 2       n         T     x     p 2   munder       R  n   1     p  2    R  n           Consider overlap of T   on   S n   1        R   R 2      Average overlap on   S n   1   Not too hard to check that the number of active T   on   S n   1 is         n   1   R 2   n   1     Now calculate    Two cases  either R 2 n dominates or the other term dominates  So p   2 n n   1