6   Tube incidence implications of Fourier restriction  Last time  LOCALLY CONSTANT PROPERTY  think of it more as  heuristic     If supp f     B 1 then   1    Imagine   f     const on unit balls    2    f         B 1       f       B 1   What if supp f     B     0    so   f     const on     1 balls    where   B             B 1       1               B 1   is approximately averaging over a 1 ball           B     is approximately averaging over     1 balls   What about supp f     B     v     Same thing happens  because e i x   something f will have  Fourier support in B     0    and taking absolute values means we don t notice the e i x   something  modulation    Returning to R P n   1     q     p       R P n   1     loc   q     p         B R   x     S     n        B R     1 on B R  supp   B R     B R   1   0     Last lecture    Can choose   B R   such that       B R       R n   B R   1   0         B R     1   1   Case 1   p   q     1  LHS of        using H lder  is  munder         I R   1   1   p   q       I R   1             1     f           B R     1 q     1 q                 2     n     q   d     d   n   p   q       I R   1   1   p   q         n     f           B R             q   d d     p   q           f     q                 B R           d     1 q   munder       1   Holder       I  R    1     1   p     q             n        n    f        q                    B  R             d    d        p     q         I  R    1     1   p     q             n    f        q           d        p     q     Case 2   p   q     1  Use P 1     2 for intuition    Imagine a function g which is approximately constant on each R   1 cube Q  Think of g as g     Q g Q   Note  Therefore   if     2     c R   1     C for all   2   I R   1        P       0     2       Q     R   1 for     2     c R   1       I R   1   1   p   q   p q   C p   q       I R   1   1   p   q       I R   1   C   p   q       R   1     t     1 g   t   t 2     2   d t d   2   p   q    Important  locally constant property means we didn t need p   q     1  like before   Make the intuition rigorous    Consider the integral                  1     f           B R                     2     n     q   d                   1       n   f               B R                       2     n         d     q   d                   1       n   f     q             B R                       2     n         d     d     Same pointwise Holder   R n   R   1   n   1     n   f     q         d     R 1     n   f     q       R   1   1   R p   q         f     q     p   q