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.