Hi,

maybe I am completely wrong, but one should be able to interpolate
since it is bounded on L^{2-\delta}. So for a \varepsilon depending on
\delta and the norm on L^{2-\delta}, one gets the norm smaller than 1.

Regards

Josef


Quoting mablung123 <david.cruzuribe@...>:

> I have a linear operator T that is bounded on L^p(w), 2-\epsilon < p
>  < 2+ \epsilon, for a fixed weight w.  I know that on L^2(w) the
> operator norm of T is less than 1.  Does it follow that for \epsilon
>  sufficiently small, the operator norm of T is less than 1 on L^p(w)?
>
> A reference would be greatly appreciated.
>
> David Cruz-Uribe, SFO
>
>



--
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.

David,

By Riesz-Thorin, the logarithm of the operator norm is convex as a function of
1/p, which means it's continuous.

Best, Philip

--- In harmonicanalysis@yahoogroups.com, "mablung123" <david.cruzuribe@...>
wrote:
>
> I have a linear operator T that is bounded on L^p(w), 2-\epsilon < p < 2+
\epsilon, for a fixed weight w.  I know that on L^2(w) the operator norm of T is
less than 1.  Does it follow that for \epsilon sufficiently small, the operator
norm of T is less than 1 on L^p(w)?
>
> A reference would be greatly appreciated.
>
> David Cruz-Uribe, SFO
>

I have a linear operator T that is bounded on L^p(w), 2-\epsilon < p < 2+
\epsilon, for a fixed weight w.  I know that on L^2(w) the operator norm of T is
less than 1.  Does it follow that for \epsilon sufficiently small, the operator
norm of T is less than 1 on L^p(w)?

A reference would be greatly appreciated.

David Cruz-Uribe, SFO

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Dear Members
can any one help me  about some qusetions from these lemma.
we fisrt have some assumptions:
Asuumptions:
  Let $\mu$  be a finite  positive regular Borel Measure on the locally compact
Hausdorff space $X$. we let $S^*(X. \mu)$  denote the elements of $L^\infty
(\mu)$ that have absolute value one $\mu$-almost every where. Let $\rho:
\Gamma\rightarrow S^*(X,\mu)$ be a group homomorphism  from  alocally  compact
abelian group $\Gamma$  such that  $\rho(\gamma)\in CB(X)$ for all $\gamma$  and
such that $\rho(\Gamma)$ seprates the points of $X$.
lemma: Under above  assumptions , let $F$  be any weak limit of a net in
$O(\rho)=\{ \rho(\gamma): \gamma\in \Gamma    \}$. then $F$ has absolute value 
one every where as an element of $L^\infty (\mu)^{**}$.
proof: Fix   a point $\omega$  in the support of $\mu$, considered  as a measure
on $\Delta(L^\infty)(\mu)$. then $|\rho(\gamma_{\alpha})|=1$
   for all $\alpha$, since evaluation of $\rho(\gamma_{\alpha})$ at $\omega$  is
given by the bounded  linear functional  $$\rho(\gamma_alpha)\rightarrow \int
\rho(\gamma_\alpha)d\delta_omega$=\prec \rho(\gamma_{\alpha}), \delta_\omega
\succ$$
which maps $L^\infty(\mu)\rightarrow C$. in  above equation the integration  is
aganist the unit point mass at $\omega\in \Delta(L^\infty)(\mu)$. and we are
identifying elements  of  $L^\infty(\mu)$ with thier Gelfand  transforms, hence
the  weak limit $F$ of elements
of $\rho(\Gamma)$ has $|F|=1$ as an element of $L^\infty(\mu)^**$.

Dear all,

does there exist a characterization of the space of harmonic functions ?

(To be clear, let us consider the Banach space of harmonic functions on the open
unit ball in R^n which have an continuous extension on the closed ball, endowed
with the sup norm.)

Thank y'all for your answers!
L.

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.

Thanks a lot.

--- In harmonicanalysis@yahoogroups.com, "Philip Gressman" <ptgressman@...>
wrote:
>
> This reply assumes that you left out a square root: (1+|m|^2)^{(n/2-1)/2}.
>
> This operator may be expressed as convolution with a kernel k(x) which has a
singularity like |x|^{-n/2-1} at the origin.  The kernel just fails to belong to
L^{2n/(n-2)} (it's in the corresponding weak space).  If it actually did belong
to this particular L^p space, the boundedness question you asked would be an
immediate consequence of Young's inequality for convolutions.  To take care of
the details, you should look at the proof of the Hardy-Littlewood-Sobolev
inequality appearing in Stein's _Harmonic Analysis_ (the proof there is for R^n,
but it goes through on the torus without any significant changes).
>
> -Philip
>
> --- In harmonicanalysis@yahoogroups.com, "lakhmau" <lakhmau@> wrote:
> >
> > Dear all,
> >
> > I would like to know why the (Bessel-like ?) operator which maps
> >
> > exp(2i.pi.m.x) |-> exp(2i.pi.m.x) / (1 + |m|^2)^{n/2 - 1}
> >
> > for any multi-index m \in \Z^n maps the space
> >
> > L^{n/(n-1)}((0,1)^n)
> >
> > into L^2((0,1)^n) ?
> >
> > Sorry, the post is not really readable...
> >
> > Thanks,
> > L.
> >
>

lakhmau wrote:
>
>
>
> Dear all,
>
> I would like to know why the (Bessel-like ?) operator which maps
>
> exp(2i.pi.m.x) |-> exp(2i.pi.m.x) / (1 + |m|^2)^{n/2 - 1}
>
> for any multi-index m \in \Z^n maps the space
>
> L^{n/(n-1)}((0,1)^n)
>
> into L^2((0,1)^n) ?
>
> Sorry, the post is not really readable...
>
> Thanks,
> L.

I have an elementary proof of a similar result at the end of the paper:
http://www.math.missouri.edu/~stephen/preprints/thin.html
You may be able to adopt this to your situation.  But really, you want
to learn the Littlewood-Paley decomposition and use that.

This reply assumes that you left out a square root: (1+|m|^2)^{(n/2-1)/2}.

This operator may be expressed as convolution with a kernel k(x) which has a
singularity like |x|^{-n/2-1} at the origin.  The kernel just fails to belong to
L^{2n/(n-2)} (it's in the corresponding weak space).  If it actually did belong
to this particular L^p space, the boundedness question you asked would be an
immediate consequence of Young's inequality for convolutions.  To take care of
the details, you should look at the proof of the Hardy-Littlewood-Sobolev
inequality appearing in Stein's _Harmonic Analysis_ (the proof there is for R^n,
but it goes through on the torus without any significant changes).

-Philip

--- In harmonicanalysis@yahoogroups.com, "lakhmau" <lakhmau@...> wrote:
>
> Dear all,
>
> I would like to know why the (Bessel-like ?) operator which maps
>
> exp(2i.pi.m.x) |-> exp(2i.pi.m.x) / (1 + |m|^2)^{n/2 - 1}
>
> for any multi-index m \in \Z^n maps the space
>
> L^{n/(n-1)}((0,1)^n)
>
> into L^2((0,1)^n) ?
>
> Sorry, the post is not really readable...
>
> Thanks,
> L.
>

Dear all,

I would like to know why the (Bessel-like ?) operator which maps

exp(2i.pi.m.x) |-> exp(2i.pi.m.x) / (1 + |m|^2)^{n/2 - 1}

for any multi-index m \in \Z^n maps the space

L^{n/(n-1)}((0,1)^n)

into L^2((0,1)^n) ?

Sorry, the post is not really readable...

Thanks,
L.

Any subset of rational numbers is a countable union of points. As all
points are congruent, you can have either m(p)=0, and then the whole
measure is 0. Or you can have a non-zero point mass a and then m(E)=a(#E),
i.e. measure is finite only for finite sets.

> Can we say that there is no countably additive invariant measure on the
> additive group of rational numbers with the subspace topology from the
> real line?
>                S.Srinivas Rau
>
>

Dear All
Thanks  a lot  Maria Roginskala. please help me with  the following
questions :
1-As I know when $G$ is a compact  group then the Haar measure $\mu$
does belong to $M(G)$. Is it true when $G$  is a locally comapct abelian
group?
2-Let  $G$ is a locally  compact  abelian  group. if we  prove that Haar
measure $\mu$  that is restircted to a relatively compact  open subset
of $G$ , say $U$ , is not discrete , can we conclude that $\mu$ is not
discrete for the group $G$?
I wish to hear from whom can give his or her  comments very soon.
Best Regards
   Fatima

Can we say that there is no countably additive invariant measure on the additive
group of rational numbers with the subspace topology from the real line?
                S.Srinivas Rau

If you have a locally compact Hausdorff space, then every point has a
local basis of compact neighbourhoods. I.e. for any point there is a
compact set which contains an open set which also contains the point. Now,
if every compact is finite, than there are finite open sets. Take such a
set E, which contains p. As the space is Hausdorff for each x in E which
is not p one can find an open set U_x which contains p and does not
contain x. Then the intersection of E and all U_x is just {p}, and as an
intersection of finitely many open sets it should be open. I.e. the
topology is discrete.

I want to remark that even for a Hausdorff topological group, unless the
unity (and thus all other points) have a countable neighbourhood basis,
the statement does not hold. I.e. one can construct a Hausdorff
topological group in which only finite sets are compact, but it is not a
discrete.

> Dear All
>  please help me with the proof of this point.
>  the group $G$ is discrete if the Haar measure $\mu$ is discrete.
>  In the proof of this point the writer has said if $G$ is not
>  discrete then it contains  a compact set that is  infinite. Is
>  there any one who can help me with the above line.
>  $G$ is also a locally compact  Haussdorff space.
>  Thanks
>  Fatima
>
>
>

Dear All
  please help me with the proof of this point.
  the group $G$ is discrete if the Haar measure $\mu$ is discrete.
  In the proof of this point the writer has said if $G$ is not
  discrete then it contains  a compact set that is  infinite. Is
  there any one who can help me with the above line.
  $G$ is also a locally compact  Haussdorff space.
  Thanks
  Fatima

Dear All
  please help me with the proof of this point.
  the group $G$ is discrete if the Haar measure $\mu$ is discrete.
  In the proof of this point the writer has said if $G$ is not
  discrete then it contains  a compact set that is  infinite. Is
  there any one who can help me with the above line.
  $G$ is also a locally compact  Haussdorff space.
  Thanks
  Fatima