A property of $l_p$ space I'd like to show the following. Let $(u_n)_{n\geq 1}$ be a sequence of elements in $l_p(\mathbb{N})$. Suppose $u_n$ is p-norm-bounded by some $m>0$ and that for every $k\geq 1$, $\lim\limits_{n\to\infty}u_n(k)=0$, then $u_n$ has limit $0$ in weak topology.
I tried consider $f\in l_p^*$ and make $f(u_n)$ go to $0$ like this:
$|f(u_n)|\leq \lVert f\rVert.\lVert u_n\rVert_p$ , but I'm stuck. Hope somebody could help!
 A: In this problem, $p$ must be greater than $1$. Here is a sketch of the proof. Let $\pi_{\leq N}$ be the coordinate projection on the subspace of $\ell_p$ spanned by $e_1, \dots, e_N$; let $\pi_{{}> N}$ be the coordinate projection on the closed subspace spanned by $e_{N+1}, e_{N+2}, \dots$. 
Consider $f\in \ell_p^*$. Note that $f = f\pi_{{}> N} + f\pi_{{} \leq N}$. We have that $\|f\pi_{{}> N} \| \to 0$ as $N\to \infty$ since
$$\|f\pi_{{}> N}\| = \left(\sum_{i=N+1}^\infty |f_i|^q\right)^{1/q}$$
(where $1/ p + 1/q = 1$).
Choose $N$ such that $\|f\pi_{{}> N} \|\leq \varepsilon$. Then, 
$$\limsup_{n\to \infty} |f(u_n)| \leq \limsup_{n\to \infty} |f\pi_{{}\leq N}(u_n)| + \limsup_{n\to \infty} |f\pi_{{}> N}(u_n)| \leq 0 + \|f\pi_{{}> N}\|\cdot m \leq \varepsilon m.$$
We get that for every $\varepsilon > 0$, $\limsup_{n\to \infty} |f(u_n)| \leq \varepsilon m$. Thus $\limsup_{n\to \infty} |f(u_n)| = 0$.
Here is a counterexample for $p=1$. Let $u_i = e_i$ and $f(u) = \sum_{i=1}^\infty u(i)$. The conditions of the problem hold but $f(u_i) = 1\not\to 0$. 
A: This isn't true for $p=1$.
For $1<p<\infty$, here is a hint:
Write $$|<f,u_n>| = \biggl|\sum_{i=1}^k f(i)u_n(i)\,\biggr| +\biggl| \sum_{i=k+1}^\infty f(i)u_n(i)\,\biggr|, $$
where $f(i)$ is the $i^{\rm th}$ coordinate of $f$ (and similarly for $u_n(i)$).
Apply Hölder's  inequality to obtain
$$
|<f,u_n>|\le
\biggl(\sum_{i=1}^k |f(i) |^q\biggr)^{1/q} \underbrace{\biggl(\sum_{i=1}^k |u_n(i) |^p\biggr)^{1/p}}_A  +\underbrace{
\biggl(\sum_{i=k+1}^\infty |f(i) |^q\biggr)^{1/q}}_B \
\underbrace{\biggl(\sum_{i=k+1}^\infty |u_n(i) |^p\biggr)^{1/p}  }_C
$$
Note: 


*

*You may  choose $k$ so large that term $B$ is as small as desired.

*Using the pointwise convergence of $(u_n)$, for a fixed $k$, you may choose $N$, so
that $n>N$ forces term $A$ to be as small as desired.

*The term $C$ is bounded since the $u_n$ are norm bounded.

