Re: [harmonic] weak conv/ strong convergence

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• Dear All, Could anyone suggest a source where I can find the difference between weak and strong conversion. Thanks and Best Regards, Ahmed Ibrahim ...
Message 1 of 4 , Mar 1, 2005
Dear All,
Could anyone suggest a source where I can find the
difference between weak and strong conversion.

Thanks and Best Regards,
Ahmed Ibrahim

--- h1l2star <h1l2star@...> wrote:

>
> Dear Members of the Harmonic Analysis
>
> I know that
> [1] u(t)\in L^{00}(I,L^{2}(R^{n})), where
> I\subset\R an interval.
> and I want to show
> [2] u(t)\in C(I,L^{2}(R^{2})).
>
> From [1] Eberlein-Smulian's Thm gives now a weak
> convergent
> subsequence \Lambda in L^{2}.
> But I want to obtain strong convergence in L^{2} for
> every sequence
> t_{k}->t for k->00.
>
> I wanted to use Lebesgues Dominated Convergence
> Theorem.
> But therefore I need convergence almost everywhere.
> And I think weak conv. doesn't imply conv. a.e. ?
>
> So I tried the following:
> Because of weak conv. in L^{2} we can choose
> \phi_{1},\phi_{2}\in C^{00}_{0}(R^{n})
> such that
> \int |(u(t_{k})-u(t))\phi_{1}|^{2}dx \leq \int
> |u(t_{k})-u(t)|^{2}dx
> \leq \int (u(t_{k})-u(t))\phi_{2}dx
> and with k->00, k\in\Lambda we obtain the strong
> conv.
>
> I used that the weak limit is equal to the strong
> limit.
> Is this always true or do I have to prove this in
> every situation?
>
> I'm very grateful for answers, remarks and hints!
>
> With best regards
>
> h1l2star
>
>
>
>

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• The notation L^{00}(I,L^{2}(R^{n})) is not clear. Please use standard notation, or better explain in words. Best, Nir. ... dx
Message 2 of 4 , Mar 28, 2005
The notation " L^{00}(I,L^{2}(R^{n})) " is not clear.

Please use standard notation, or better explain in words.

Best, Nir.

--- In harmonicanalysis@yahoogroups.com, "h1l2star" <h1l2star@y...>
wrote:
>
> Dear Members of the Harmonic Analysis
>
> I know that
> [1] u(t)\in L^{00}(I,L^{2}(R^{n})), where I\subset\R an interval.
> and I want to show
> [2] u(t)\in C(I,L^{2}(R^{2})).
>
> From [1] Eberlein-Smulian's Thm gives now a weak convergent
> subsequence \Lambda in L^{2}.
> But I want to obtain strong convergence in L^{2} for every sequence
> t_{k}->t for k->00.
>
> I wanted to use Lebesgues Dominated Convergence Theorem.
> But therefore I need convergence almost everywhere.
> And I think weak conv. doesn't imply conv. a.e. ?
>
> So I tried the following:
> Because of weak conv. in L^{2} we can choose
> \phi_{1},\phi_{2}\in C^{00}_{0}(R^{n})
> such that
> \int |(u(t_{k})-u(t))\phi_{1}|^{2}dx \leq \int |u(t_{k})-u(t)|^{2}
dx
> \leq \int (u(t_{k})-u(t))\phi_{2}dx
> and with k->00, k\in\Lambda we obtain the strong conv.
>
> I used that the weak limit is equal to the strong limit.
> Is this always true or do I have to prove this in every situation?
>
> I'm very grateful for answers, remarks and hints!
>
> With best regards
>
> h1l2star
• Try these books 1. Functional Analysis: An Introduction / Yuli Eidelman, Vitali Milman and Antonis Tsolomitis 2. Introductory real analysis / A.N. Kolmogorov
Message 3 of 4 , Mar 28, 2005
Try these books

1. Functional Analysis: An Introduction / Yuli Eidelman, Vitali
Milman and Antonis Tsolomitis
2. Introductory real analysis / A.N. Kolmogorov and S.V. Fomin

But actually these notions are very simple, so try

Best, Nir.

--- In harmonicanalysis@yahoogroups.com, ahmed ibrahim
<asony11@y...> wrote:
> Dear All,
> Could anyone suggest a source where I can find the
> difference between weak and strong conversion.
>
> Thanks and Best Regards,
> Ahmed Ibrahim
>
> --- h1l2star <h1l2star@y...> wrote:
>
> >
> > Dear Members of the Harmonic Analysis
> >
> > I know that
> > [1] u(t)\in L^{00}(I,L^{2}(R^{n})), where
> > I\subset\R an interval.
> > and I want to show
> > [2] u(t)\in C(I,L^{2}(R^{2})).
> >
> > From [1] Eberlein-Smulian's Thm gives now a weak
> > convergent
> > subsequence \Lambda in L^{2}.
> > But I want to obtain strong convergence in L^{2} for
> > every sequence
> > t_{k}->t for k->00.
> >
> > I wanted to use Lebesgues Dominated Convergence
> > Theorem.
> > But therefore I need convergence almost everywhere.
> > And I think weak conv. doesn't imply conv. a.e. ?
> >
> > So I tried the following:
> > Because of weak conv. in L^{2} we can choose
> > \phi_{1},\phi_{2}\in C^{00}_{0}(R^{n})
> > such that
> > \int |(u(t_{k})-u(t))\phi_{1}|^{2}dx \leq \int
> > |u(t_{k})-u(t)|^{2}dx
> > \leq \int (u(t_{k})-u(t))\phi_{2}dx
> > and with k->00, k\in\Lambda we obtain the strong
> > conv.
> >
> > I used that the weak limit is equal to the strong
> > limit.
> > Is this always true or do I have to prove this in
> > every situation?
> >
> > I'm very grateful for answers, remarks and hints!
> >
> > With best regards
> >
> > h1l2star
> >
> >
> >
> >
>
>
>
>
> __________________________________
> Do you Yahoo!?
> Yahoo! Mail - Find what you need with new enhanced search.
> http://info.mail.yahoo.com/mail_250
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