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Let 1 < p < ∞. Prove that a sequence $x^{(n)}_n⊂ l^p$ is weak convergent to $x ∈ l^p$ if and only if $x^{(n)}_n$ is bounded and pointwise convergent, i.e. $supn∈N ||x^{(n)}||_p< ∞ \;and \;x^{(n)}_k n→∞ −−−−→ x_k$for every k ∈ N. Is this equivalence still true for p = 1 or p = ∞???? Do we start with letting X be a general Banach space and X∗ its dual?? Do we assume xn weakly converges to x and also x is bounded? Thanks for any help!

  • Working in general Banach spaces won't help; this property is special to $\ell^p$ spaces. (The concept of "pointwise convergent" doesn't even make sense in a general Banach space.) However, you will want to remember some general Banach space facts, like the uniform boundedness principle. – Nate Eldredge Dec 19 '22 at 21:52
  • I'm not sure what you mean by "$x$ is bounded". But for instance, it should be clear that any $x \in \ell^p$ is bounded as a sequence of real (or complex) numbers, i.e. $\sup_k |x_k| < \infty$, simply because $|x_k| \le |x|_p$ for every $k$. – Nate Eldredge Dec 19 '22 at 21:54
  • What about if p is 1 or inf how and why doe it change the weak convergeants –  Dec 21 '22 at 00:05
  • In $\ell^1$, for sequences, weak convergence is equivalent to convergence in norm; this is Schur's theorem. In $\ell^\infty$ things are more difficult because the dual $(\ell^\infty)^*$ is complicated to describe. – Nate Eldredge Dec 21 '22 at 18:25
  • But for quick counterexamples: in $\ell^1$, let $x^{(n)}k = \delta{n,k}$, i.e. $x^{(n)}$ has 1 as its $n$th element and 0 everywhere else. Every $x^{(n)}$ has norm 1, and $x^{(n)}_k \to 0$ for every $k$, so the sequence converges pointwise to 0. However the linear functional $f(x) = \sum_k x_k$ is continuous and satisfies $f(x^{(n)})=1$ for all $n$. – Nate Eldredge Dec 21 '22 at 18:27
  • In $\ell^\infty$, let $x^{(n)}k = 1$ for all $k \ge n$ and 0 otherwise. This converges pointwise to 0 and has constant norm 1. Consider the closed subspace $C$ of $\ell^\infty$ consisting of sequences $x$ for which $\lim{k \to \infty} x_k$ exists, and consider the continuous linear functional $f_0$ on $C$ defined by $f(x) = \lim_{k \to \infty} x_k$. Note $x^{(n)} \in C$ and $f_0(x^{(n)}) = 1$ for all $n$. Use Hahn-Banach to extend $f_0$ to a continuous linear functional $f$ on $\ell^\infty$. Then we still have $f(x^{(n)}) = 1$ for every $n$. – Nate Eldredge Dec 21 '22 at 18:31

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