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A question on operators mapping L^p to L^q

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  • andredelaire
    Dear all, As you know, the linear bounded operators mapping L^p(R^N) to L^q(R^N) (that commute with translations) are given by a convolution of a tempered
    Message 1 of 1 , Jun 16, 2009
      Dear all,
      As you know, the linear bounded operators
      mapping L^p(R^N) to L^q(R^N) (that commute
      with translations) are given by a convolution
      of a tempered distribution.

      A classical result says that
      the possible values of (p,q)
      satisfy that the set (1/p,1/q)
      is a convex (in the square [0,1]x[0,1])

      For example (the convolution with) 1/abs(x)^(N+a),
      with 0<a<N, maps L^p(R^N) to L^q(R^N) with
      1/p-1/q=a/N (1<p<q<infinity). In this case
      the convex is a line.

      If we take T as function in L^1 and
      that also belongs to L^2, then (by
      Young's inequality)the
      convex set of (1/p,1/q) s.t. the convolution
      with T maps L^p(R^N) to L^q(R^N) is a
      quadrilateral in the square [0,1]x[0,1])

      Now my problem: does somebody know
      an EXAMPLE of a distribution s.t.
      (by the convolution) maps L^p(R^N) to L^q(R^N)
      and the set of (1/p,1/q) be a TRIANGLE
      (in the square [0,1]x[0,1])???
      In particular, I'd like that one of the vertices
      of the triangle was in (1/2,1/2).

      Sorry for the long introduction,
      but I tried to be clearly as i could.

      Thanks for your answers.
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