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Let $1\leq p < \infty$. Prove that if any sequence $(x_n)_{n \in \mathbb{N}} \subset l^p$ which converges weakly to $0 \in l^p$ converges in norm to $0$, then $p=1$.

I started with assuming that $p>1$ then I found a sequence $(e_n)_n$ the standard basis in $l^p$ for which this isnt true. ($e_n \to 0$ (weak) but the norm is $1$).

Can someone help me with the case $p=1$?

Olba12
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  • The case $p=1$ is trivial. What you have to show is that $p=1$, and this is fulfilled. – PhoemueX Apr 15 '17 at 20:29
  • How is the case $p=1$ trivial? @PhoemueX – Olba12 Apr 15 '17 at 20:30
  • Your task is the following: "Assume XYZ, then show $p=1$." But in the case $p=1$, it is trivial that $p=1$. – PhoemueX Apr 15 '17 at 20:32
  • Yes, but I need help showing that it is fufilled when p=1. @PhoemueX – Olba12 Apr 15 '17 at 20:33
  • This is not what you stated as your task, but see here: https://math.stackexchange.com/questions/42609/strong-and-weak-convergence-in-ell1. EDIT: Clarification: Your task (as you posted it) does NOT say that you need to show that XYZ holds if $p=1$. It just says that you need to show that XYZ can only ever be true if $p=1$. – PhoemueX Apr 15 '17 at 20:34
  • Yes, that is why I started with $p>1$ To show that it couldnt be true. Why cant I then assume p=1 and then show that it is in fact true? @PhoemueX – Olba12 Apr 15 '17 at 20:40
  • This is what you want to do, and I think that you will find PhoemueX's link helpful. Some might find it unclear from your post that you already showed the property does not hold whenever p>1. – M A Pelto Apr 15 '17 at 22:07

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