Weakly convergence characterization for $l^p$ For $1 \leq p < \infty$, let $\{x_n\}$ be a bounded sequence in $l^p$ and $x$ belong in $l^p$. Show that $\{ x_n \} \rightharpoonup x$ in $l^p$ if and only if it converges componentwise, that is, for each index $k$,
$$\lim_{n \to \infty} x_n^k =x^k  \text{ where } x_n = ( x_n^1, x_n^2, ...) \text{ and } x= (x^1, x^2...).$$
I try like this: but I think this is true only if $x_n$ converge strongly not weakly. Any solution for this problem?
$"\Rightarrow"$
Let $(x^n)=(x_1^n, x_2^n,..., x_j^n...) \rightharpoonup x=(x_1, x_2,..., x_j,..)$ a bounded sequence in $l^p$
\begin{align*}
\Rightarrow & ||x^n - x||_p < \epsilon\\
& \sum\limits_{j=1}^{\infty} |x_j^n - x_j|^p < \epsilon
\end{align*}
Fix $j :$ $$ |x_j^n - x_j|^p < \sum\limits_{j=1}^{\infty} |x_j^n - x_j|^p < \epsilon$$
$$\Rightarrow |x_j^n - x_j|^p <\epsilon$$
$$\Rightarrow \lim\limits_{n \to \infty} x_j^n = x_j \ \forall j$$
$"\Leftarrow"$ need help
 A: Your reasoning is wrong. Your assumption is that $\{x_n\}_{n\in\mathbb N}$ converges weakly to $x$ and not in the norm of the space $l^p.$ 
Necessity: if $\{x_n\}_{n\in\mathbb N}$ converges weakly to $x$, then by the standard duality $(l^p)^*\cong l^q$ where $1/p+1/q=1.$ In this case 
$e_k(x_n)\underset{n\to \infty}{\longrightarrow} e_k(x)$ yields componentwise convergence. 
Sufficiency: Let us write $\langle y,x\rangle=y(x)$ for $x\in l^p$ and $y\in l^q.$ Let us choose $y\in l^q$ and $\epsilon>0$ arbitrary. Then there exists (why) $k\in \mathbb N$ such that $\|y-y^{(k)}\|_q<\frac{\epsilon}{2M}$ where $y^{(k)}$ is the projection of $y$ to the subspace spanned by vectors $e_1,\ldots,e_k$ and $M=\sup_{n\in\mathbb N}\|x-x_n\|_p.$ 
\begin{align}
|\langle y,x-x_n\rangle| \leq |\langle y-y^{(k)},x-x_n\rangle|+|\langle y^{(k)},x-x_n\rangle|.
\end{align}
By applying Holder inequality we obtain 
$$
|\langle y-y^{(k)},x-x_n\rangle|\leq \|y-
y^{(k)}\|_q \cdot \|x-x_n\|_p<\epsilon.$$ Since $\{x_n\}_{n\in\mathbb N}$ converges to $x$ componentwise and since $y^{(k)}$ has only finitely many nonzero components, there exists $k'\in \mathbb N$ such that 
$|\langle y^{(k)},x-x_n\rangle|<\frac{\epsilon}{2}$ for all $n\geq k'.$
Remark: you do not need to assume that $\{x_n\}_{n\in\mathbb N}$ is bounded since every weakly convergent sequence is always bounded as a quick consequence of the principle of the uniform boudnedness. 
A: Here is a proof for the characterization for the weak convergence in the case $p\in (1,+\infty)$. 
Now, the result you are trying to prove (that the sequence $(x_n)_n$ converges strongly to $x\in \ell^p$ if and only if it converges pointwise and is bounded) is false for $p>1$. The counter-example in this case is the sequence $(e_n)_n$ such that for all $k,n$, 
$$
e_n^k=
\begin{cases}
1 & k=n\\
0 & k\neq n
\end{cases}
$$ 
You can see this sequence is in $\ell^p$, is bounded (by 2), and it converges pointwise to $0\in \ell^p$, but does not converge strongly to $0$ in $\ell^p$.
A: By def'n, $(x(n))_n$ converges weakly to $x$ iff $(f(x(n))_n$ converges to $f(x)$ for every $f\in l_p^*.$ To prove that a bounded sequence $x(n)_n=(<x(n)_i>_i)_n$ converges weakly to $<x_i>_i$ iff $(x(n)_i)_n$ converges to $x_i$ for each $i$ ,we have:(1) Necessity.For each $i$ let $f_i(<y_i>_i)=y_i$ for all $<y_i>_i\in l_p.$ Then $x_i=f_i(x)=\lim_{n \to \infty} f_i (x(n))=x(n)_i.$ (2) Sufficiency. For $m\in N$: For  $y=<y_i>_i\in l_p$ let $$(j\leq m\implies(y|_m)_j=x_j)\;\land \;(j>m \implies (y|m)_j=0.$$ $$\text { For } (f\in l_p^*\; \land \;  y\in l_p) \text { let }\; f|_m(y)=f(y|m).$$  We use the facts that $$(a)\quad \forall x\in l_p \;(\lim_{m\to \infty}\|x-x|m\|=0;$$ $$ (b)\quad \forall f\in  l_p^*\;(\lim_{m\to \infty}\|f|_m-f\|=0.$$ Now take any $f\in l_p^*$:  Given $e>0$ take $m$  large enough that  $\|x-x|m\|<e$ and $\|f-f|m\|<e.$ Now take $n$ large enough that $n'>n\implies  \forall j\leq m\; (|x_i-x(n)_j|<e/m ).$ Then $n'>n\implies$ $$ |f(x)-f(x(n'))|=|f(x|_m-x(n')|_m)+ (f-f|m) (x-x(n')|\leq$$ $$\leq e \|f\| +\|f-f|_m\|\cdot (\|x\|+\|x(n')\|)\leq$$ $$\leq e\|f\|+\|f-f|m\|\cdot(\|x\|+K)$$ where $ K=\sup_{i\in N}\|x(i)\|.$ Letting $n \to \infty$ we have $\lim_{n\to \infty}$      $\sup_{n'>n}$ $ |f(x)-f(x(n')|\leq e$ $ \|f\|+\|f-f|_m\| \cdot  K.$ Now letting $m\to \infty,$ by (b) we have $\lim_{n\to  \infty} \sup_{n'>n}|f(x)-f(x(n))|\leq e.$  Since $e>0$ is arbitrary, we are done..... Note that the condition $K<\infty$ is necessary. Note that we also used $f|_m(x)=f(x|_m)$ which may help to explain one of the lines above.  
