2   Exponential sums in L p  Recall  studying f   x           R e       x    When R does not have  linear  structure  expect   f   x         R   1 2   sqrt cancellation   in an appropriate sense  in L p   more  than when R is structured   Linear R f   x       n   1 N e   n x   vs convex R g   x       n   1 N e   n 2 x     Have   Have   f   x       N on   0   c N   1    and   g   x       N on   0   c N   2     L 2 does not  distinguish f   g  L   does not distinguish  However  2   p     does   What about 1   p   2  We don t usually study this range because the estimates tend to be trivial   not interesting   Focus on 2   p      Preduct size of   Square root cancellation lower bound      N p   2       0   1     f   p   Constant integral lower bound          0   1     f   p       0   c N   1     f   p   N p N   1     0   1     g   p       0   c N   2     g   p   N p N   2         f   p   N p   2   N p   1      g   p   N p   2   N p   2   Note that for the f bound  N p   1 is bigger than N p   1  as long as p   2    so the constant integral dominates   For g  if 2   p   4  the square root cancellation dominates  but for p   4 the constant integral takes over   Assuming   Theorem 2 1    p   2   b n     Then   Proof  Consider    n   1 N b n e   n x    b n              0   1       n b n e   n x     p d x       n b n e   n x       p   2     0   1       n b n e   n x     2 d x     N 1 2   b n   2   p   2   b n   2 2 CS   N p 2   1   b n   2 p       Note that this is sharp when b n   1     n   1 N b n e   n 2 x    Focus on p   4   The integral vanishes unless n 1 2   n 3 2   n 4 2   n 2 2   Number Theory lemma  If m      then   Follows from unique prime factorisation   Warning   The above lemma is false   See correction later    For fixed n 1   n 3   Hence         0   1       n b n e   n 2 x     4 d x     n 1   n 3   b n 1 b n 3         n 2   n 4     S n 1   n 3 b n 2 b n 4       log N     b n   2 4     We will now use   to mean   up to powers of log N   2   p   4  1 p     2   1     4        0   1    h     n b n e   n 2 x      h   p     h   2     h   4 1         b n   2     b n   2 1         b n   2   p   4   Assuming   Theorem 2 2    2   p Then    Sharp by b n   1   Positive take away  estimates are sharp  proofs are elementary  Easy to think of sharp examples   Number Theory counting idea shows          0   1   2   u   x   t     6 d x d t     b n   2 6 u   x   t         n     N n     b n e   n x   n 2 t       0   1   3   u   x   t     4 d x d t     b n   2 4 u   x   t         n     N n     2 b n e   n   x     n   2 t       Sharp  6   4   p crit   Strichartz estimate for periodic Schr dinger equation  observed by Bourgain in 1990s   Negatives  on   3   p crit   1 0 3  but this technique can only work on even integer values of p     1    2 only sharp Strichartz estimates per Schr dinger until 2015   2015  Bourgain Demeter proved   l 2   L p   sharp decoupling estimate  Gives sharp Strichartz estimate for Schr dinger in   d for all d   Proved earlier     where n   N means N   n   2 N    Conjecture    a n     0   1    a n   1   a n   1 N  a n   2   a n   1     a n   1   a n     1 N 2   Example    a n       n 3 N 3   n   N 2 N