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For 1 < p < oo l^p

  1. Oct 25, 2008 #1
    For 1 < p < oo it is true that:
    If we have a sequence ( x_n )_(n >= 1) in l^p that converges weakly to zero then that implies the x_n are uniformly bounded and that ((x^{m})_n) -> 0 in C (as n -> oo) for each fixed m.
    (where l^p is the space of p-summable sequences of complex numbers, and I wrote x_n = ((x^{m})_n )_(m >= 1) , couldn't think of a better notation...)

    Is the converse true? Why would that be? It doesn't look too obvious... at least not for me.
  2. jcsd
  3. Oct 25, 2008 #2


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    Re: l^p

    Provided I've read your post correctly, then yes, the converse is true.

    So basically you want to show that [itex]|f(x_n) - f(0)| \to 0[/itex] for all [itex]f \in (\ell^p)^\ast = \ell^q[/itex] given pointwise convergence and uniform boundedness. First consider the space [itex]\ell[/itex] of finite sequences. For every [itex]f \in \ell^\ast[/itex], we have that [itex]\lim_n f(x_n) = 0[/itex] (why?). Now use the fact that [itex]\ell[/itex] is dense in all the [itex]\ell^p[/itex] spaces.
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