Let $1Proof.
(⇒) By Proposition 4.3 $(x^{(n)})$ converges weakly implies it is bounded.
The standard basis vectors $e_k$ belong to every $l^p$, and hence $e_k ∈ l^q=(l^p)^*$ where $q=\frac{p}{p-1}$.
If $x^{(n)}=(x^{(n)}_k)_{k ∈ ℕ}$ and $x=(x_k)_{k ∈ ℕ}$ and $x^{(n)} ⇀ x$, then for each $j$
\[
x^{(n)}_j=⟨x^{(n)},e_j⟩→⟨x,e_j⟩=x_j.
\]
(⇐) For all $y∈l^q$ and $ϵ>0$, let $y^{(k)}$ be the sequence $y$ truncated at position $k$.
Since $y^{(k)}→y$, ∃$k∈ ℕ$ such that $\|y-y^{(k)}\|_q\sup_{n∈ℕ}\|x-x^{(n)}\|_p<\frac{ϵ}{2}.$
By Hölder's inequality
$$
|⟨y-y^{(k)},x-x^{(n)}⟩|≤ \|y-y^{(k)}\|_q⋅\|x-x^{(n)}\|_p<\frac{ϵ}2.$$
Since $y^{(k)}_j=0$ for $j>k$ and $x^{(n)}_j→x_j\ ∀j≤k$, there is $N∈ ℕ$ such that $|⟨y^{(k)},x-x^{(n)}⟩|<\frac{ϵ}{2}$ for $n≥N$.
\[
|⟨y,x-x^{(n)}⟩|≤|⟨y-y^{(k)},x-x^{(n)}⟩|+|⟨y^{(k)},x-x^{(n)}⟩|<ϵ
\]
Since ϵ is arbitrary, $⟨y,x-x^{(n)}⟩→0$. Since $y$ is arbitrary, $x^{(n)}⇀x$.
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