Weakly Convergent l^p Sequences: Boundedness and Convergence in C

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SUMMARY

In the context of weakly convergent sequences in the space l^p, it is established that for 1 < p < ∞, a sequence (x_n) converging weakly to zero implies that the sequence is uniformly bounded and converges to zero in C for each fixed m. The discussion also confirms that the converse is true, demonstrating that pointwise convergence and uniform boundedness lead to the conclusion that |f(x_n) - f(0)| approaches zero for all f in the dual space (l^p)∗ = l^q. This conclusion relies on the density of the space l in all l^p spaces.

PREREQUISITES
  • Understanding of weak convergence in functional analysis
  • Familiarity with the properties of l^p spaces
  • Knowledge of dual spaces and their significance in analysis
  • Basic concepts of pointwise convergence and uniform boundedness
NEXT STEPS
  • Study the properties of weak convergence in l^p spaces
  • Explore the implications of the Banach-Alaoglu theorem
  • Learn about the relationship between l^p and l^q spaces
  • Investigate applications of weak convergence in functional analysis
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Mathematicians, particularly those specializing in functional analysis, graduate students studying sequence spaces, and researchers exploring convergence properties in l^p spaces.

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

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

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