I would like to know, why:If X is a subspace of L p ( G ) such that X ¯ ≠ L p ( G ), then there exists g ∈ L q ( G ), 1 p + 1 q = 1 such that ∫ G f ( x ) g ( x ) d x = 0 ; ∀ f ∈ X ¯ .Where G is a locally compact group with Haar measure d x.

Question

I would like to know, why:If   X is a subspace of       L    p    (  G  ) such that       X    ¯    ≠      L    p    (  G  ),  then there exists   g  ∈      L    q    (  G  ),       1    p    +      1    q    =  1 such that      ∫    G    f  (  x  )    g  (  x  )    d  x  =  0  ;    ∀  f  ∈      X    ¯    .Where   G is a locally compact group with Haar measure   d  x.

svirajueh · Accepted Answer

Since       X    ¯   is a proper closed subspace of the Banach space       L    p    (  G  ), the quotient space       L    p    (  G  )      /        X    ¯   is a nonzero Banach space. So there is some nonzero linear functional   ϕ on       L    p    (  G  )      /        X    ¯  , whose composition with the quotient map is a nonzero linear functional on       L    p    (  G  ) that vanishes on       X    ¯  . Finally that functional corresponds to some   g  ∈      L    q    (  G  ) since       L    q   is the dual of       L    p  .

I would like to know, why: If X is a subspace of L p </msup> ( G )

Answered question

Answer & Explanation

New Questions in Pre-Algebra