5   Equivalent Versions of Fourier Restriction  Searching for   p   q   for which    1    R     0   1   n   1     0     2        B R 2         T     p 2   continuum incidence geometry problem     integral equals     B R   B R 2   B R         T     p 2    fractal geometry     Reminder  R S n   1   p   q   means   Conjecture   Restriction Conjecture   R S n   1   p   q    if and only if n   1   n   1 q   n   1 p and p   2 n n   1   First proved for S 1     2 by Fefferman  1970  and Zygmund  1974    Special things happen in   2  classical harmonic analysis techniques apply   Same conjecture for R P n   1   p   q    where     f     L q   P n   1   q             1   f               2     q d     For n   3  open and active   Restriction theory can be used to deduce continuum incidence geometry estimates   Surprisingly  we can go the other way too  very recent progress  whereas the above direction has been well known since at least the 9 0s    Equivalent formulations of R P n   1   p   q    Dual version is called  Fourier extension      1 q   1 q     1   The integral equals   Math input error     Math input error     Call the last inequality R P n   1     q     p       Local  dual version   allows us to work with functions  F T   For any R   1  any B R     n  we have   for all f   S     n   with supp f     N R   1   P n   1     Call this R P n   1     loc   q     p       R P n   1     q     p       R p n   1     loc   q     p      First thing bounds E g  while secound thing bounds f when we have supp f     N R   1   P n   1     Let f   S     n    supp f     N R   1   P n   1    Try to express f in trems of ext  op   Math input error     Math input error            using  R  P  n   1       q       p        Goal is to bound this last expression by  Lucy case  p   q     1  i e  1   q   p    Then   s   1  actually   since h approximately constant on A   R P n   1     q     p       R P n   1     loc   q     p     continued   Spatial  x   f   x     E g   x    Frequency     g                 2     g          supp f     N R   1   P n   1     Aiming for   Lucky case  p   q     1   R     p     1     R   1   1   p   q     R   p   q  1   1 q   1 q     Unlucky case  p   q     1  H lder s inequality goes in the wrong direction    for all s   1  h   0   This is equality if h is constant on A   PAUSE THIS   Useful Harmonic Analysis tool  The locally constant property   Convolution  Let f   g   S     n    Define f   g   S     n   by   See Young s convolution   when 1   1 r   1 p   1 q   Example  g     B  B is the unit ball in   n    RHS is  average value of f on B   x       f     B is approximately constant on balls of radius   1    Support property   supp f   A  supp g   B  supp f   g   A   B     a   b   a   A   b   B     Convolution and Fourier Transform   f g     f     g     Locally constant property   Let f   S     n    supp f     B 1   0    Then for any unit ball B       n and any x   B     1   1     y   2   m is   1 on    y     1   Digesting f   0   For any unit interval I       n  x   I     Suppose this    LHS has to be constant   Lemma   Locally constant property   Assuming that   f   S     n    supp f     B 1   0    Then  for any unit ball B 1     n and any x   B     The proof of this fact is more important than the statement   we will be using the strategy in future   Proof of locally constant property   Let f   S     n    supp f     B 1   0    Let     C c       n   such that     1 on B 1   0    supp     B 2   0     By Fourier inversion    Let x 0   x   B 1  unit ball in   n    Math input error     Math input error     Returning to R P n   1     q     p       R P n   1     loc   q     p      where   B R   S     n   satisfies   B R   1 on B R  supp   B R     B R   1   0     supported in 2 R   1 neighbourhood of P n   1   Repeat steps of proof        B R   f   x     p   d x   R     p     1     I R   1               1   h   n           q   d       p   q   d   n