# For 1 < p < oo l^p

1. Oct 25, 2008

### e12514

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. Oct 25, 2008

### morphism

Re: l^p

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.