Exercise 3.19 in Brezis' Functional Analysis

Question

Exercise 3.19:

Let $E = \ell^p$ and $F = \ell^q$ with $1 < p < \infty$ and $1 < q < \infty.$ Let $a:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $$|a(t)| \leq C|t|^{\frac{p}{q}},\quad \forall t \in \mathbb{R}.$$ Given $$x = (x_1,x_2,...,x_i,...)\in\ell^p,$$ set $$Ax = (a(x_1),a(x_2),...,a(x_i),...).$$

1. Prove that $Ax\in\ell^q$ and that the map $x\mapsto Ax$ is continuous from $\ell^p$ (strong) into $\ell^q$ (strong).

2. Prove that if $(x^n)$ is sequence in $\ell^p$ such that $x^n \rightharpoonup x$ in $\sigma(\ell^p,\ell^{p'})$ then $Ax^n \rightharpoonup Ax$ in $\sigma(\ell^p,\ell^{p'}).$

3. Deduce that A is continuous from $B_E$ equipped with $\sigma(E,E^*)$ into $F$ equipped with $\sigma(F,F^*).$

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I wrote down my attempt by looking at the answers in the back of the book. But in my attempt and in the answer at the back of the book I don't understand two points marked with bold why respectively, can you please help me to check below where my idea of proof is wrong and further, how can I solve my two why?

My Attempt:

1. First prove that $\forall x \in \ell^p$ has $Ax \in \ell^q$.Apparently. $$(\sum\limits_{k=1}^{\infty}|a(x_k)|^q)^{\frac{1}{q}}\leq |C|(\sum\limits_{k=1}^{\infty}|x_k|^p)^{\frac{1}{q}}\leq |C|(\sum\limits_{k=1}^{\infty}|x_k|^p)^{\frac{p}{q}}<\infty$$

Prove again that if there is $x^n\to x$(strong) then there is $Ax^n\to Ax$ i.e. there is $\sum\limits_{k=1}^{\infty}|x_k^n-x_k|^p \to 0$ Proof $\sum\limits_{k=1}^{\infty}|a(x_k^n)-a(x_k)|^q\to 0.$

Clearly there is $x_k^n\to x_k\forall k.$ Since $a$ is a continuous function on $\mathbb{R} $,so $a(x_k^n)\to a(x_k).$ Therefore, it is only necessary to show that $\exists N\in \mathbb{N} $such that $\sum\limits_{k=N}^{\infty}|a(x_k^n)-a(x_k)|^q\to 0$

Since $x^n \in \ell^p, \exists N_1\in \mathbb{N}$ such that (why)$$\sum\limits_{k=N_1}^{\infty}|x_k^n|^p\to 0,\quad\forall n,$$ and $x\in \ell^p, \exists N_2\in \mathbb{N}$ such that $$\sum\limits_{k=N_2}^{\infty}|x_k|^p\to 0$$

Therefore, take $N=\max{N_1,N_2},$ from Minkowski's inequality we have $$ (\sum\limits_{k=N}^{\infty} |a(x_k^n) - a(x_k)|^q)^{\frac{1}{q}} \leq(\sum\limits_{k=N}^{\infty}|a(x_k^n)|^q)^{\frac{1}{q}}+(\sum\limits_{k=N}^{\infty}|a(x_k)|^q)^{\frac{1}{q}}\leq |C|[(\sum\limits_{k=N}^{\infty}|a(x_k^n)|^p)^{\frac{1}{q}}+(\sum\limits_{k=N}^{\infty}|a(x_k)|^p)^{\frac{1}{q}}]\to 0 $$ so $Ax^n\to Ax$

2. From Exercise 3.17 if $x^n\to x\in \sigma(\ell^p,\ell^{p'})$, we have

(a)$x^n$ is bounded in $\ell^p,$

(b)$x_i^n\to x_i(n\to\infty)$ for every $i$, where $x^n = (x_1^n,x_2^n,...,x_i^n,...)$ and $x = (x_1,x_2,...,x_i,...).$

So to prove that $Ax^n\rightharpoonup Ax,$ we take $\forall y\in\ell^p,$ have $y = (y_1,y_2,... ,y_i,...) ,$ and $$\sum\limits_{k=1}^{\infty}|y_k|^p <\infty,$$ we have $$|y(Ax^n)-y(Ax)| = |\sum\limits_{k=1}^{\infty}y_k[a(x_i^n)-a(x_i)]|\leq (\sum\limits_{k=1}^{\infty}|y_k|^p)^{\frac{1}{p}}(\sum\limits_{k=1}^{\infty}|a(x_i^n)-a(x)|^q)^{\frac{1}{q}}$$ Since the proof of 1 uses only coordinate convergence i.e., the proof, here similarly.

3. This part of the attempt copied the answer, but I don't know why.

Theorem 3.29.

Let $E$ be a Banach space such that $E^⋆$ is separable. Then $B_E$ is metrizable in the weak topology $\sigma(E, E^⋆).$

Conversely, if $B_E$ is metrizable in $\sigma(E, E^⋆),$ then $E^⋆$ is separable.

The space $B_E$ is metrizable for the topology $\sigma(E, E^⋆)$ (by Theorem 3.29). Thus it suffices to check the continuity of $A$ on sequences