$\{x^n\}$ is weak convergent iff $\{||x^n||\}$ is bounded and for each $m$ we have that $\displaystyle\lim\limits_{n\rightarrow\infty} x_m^n=x_m$

Question

Let $x=\{x_m\}$ and $x^n=\{x_m^n\}$ for each $n\geq 1$. Let $1<p<\infty$ and suppose that $x\in l_p$ and $x^n\in l_p$ for each $n\geq 1$

$\{x^n\}$ is weak convergent iff $\{||x^n||\}$ is bounded and for each $m$ we have that $\displaystyle\lim\limits_{n\rightarrow\infty} x_m^n=x_m$

My attempt:
$(\implies)$ For banach-Steinhaus we have that as $\{x^n\}$ is weak convergent this implies $\{||x^n||\}$ is bounded.

Moreover, we need prove that $\displaystyle\lim\limits_{n\rightarrow\infty} x_m^n=x_m$.

Let $\epsilon>0$, $\exists N>0\::n>N$ then $||x_m^n-x_m||_p<||x_m||_p+||x_m||_p$ by Minkowski inequality.

Here, i'm stuck

Can someone help me?

($\Longleftarrow$) As convergence in norm implies weak convergence then we have that $\{x^n\}$ is weak convergent

is it correct?

Answer 1

"$\Rightarrow$" The Banach-Steinhaus part is fine - though maybe be more explicit about it. Regarding the convergence: Recall the definition of weak convergence. Since we are considering $\ell_p$, Riesz-Fischer tells us that

$$ \forall y \in \ell_q: \vert \langle y,x^n \rangle - \langle y, x \rangle \vert = \vert \sum_{i = 1}^{\infty} y_i x^n_i - \sum_{i = 1}^{\infty} y_i x_i \vert \rightarrow 0, ~~~ n \rightarrow \infty$$

Hence in particular this is true for $\delta^m_i \in \ell_q$, which is $1$ precisely when $m = i$ and $0$ otherwise. Then we have

$$ \vert \langle \delta^m_i,x^n \rangle - \langle \delta^m_i, x \rangle \vert = \vert \sum_{i = 1}^{\infty} \delta^m_i x^n_i - \sum_{i = 1}^{\infty} \delta^m_i x_i \vert = \vert x^n_m - x_m \vert \rightarrow 0, ~~~ n \rightarrow \infty ~~.$$

What you wrote down will not work as you need to assume $m \in \mathbb{N}$ i.e. $\Vert x_m \Vert_p + \Vert x_m \Vert_p$ is constant. Also, note that $x_m$ is an element of $\mathbb{R}$ (or whatever your values your sequences take)

"$\Leftarrow$" None of your assumptions give strong convergence (i.e. convergence in norm)- what you have is pointwise convergence.

A more general proof of this direction can be found here, but I'll give a brief argument:

Assume $\Vert x^n \Vert_p < M$ for some $M>0$. Let $y \in \ell_q$ and $\varepsilon > 0$ be arbitrary. Then since $\{\delta^m_i : m \in \mathbb{N}\}$ is dense in $\ell^p$, we may choose $g = \sum_{i \in \{i_1, \ldots, i_k\}} \delta^m_i$ some finite linear combination of the deltas s.t. $\Vert y - g \Vert < \frac{\varepsilon}{3M}$ and $\Vert y - g \Vert < \frac{\varepsilon}{3 \Vert x \Vert}$.

Now choose $N \in \mathbb{N}$ s.t. $\forall n > N$ we have $\vert x^n_m - x _m\Vert < \frac{\varepsilon}{3n}$. Then for any $n > N$ we have

$$ \begin{align} \vert y(x^n) - y(x) \vert & \leq \vert y(x^n) - g(x^n) \vert + \vert g(x^n) - g(x) \vert + \vert g(x) - y(x) \vert \\ & \leq \Vert y - g \Vert_q \Vert x^n \Vert_p + \vert \sum_{m \in \{m_1, \ldots, m_k\}} x^n_{m_k} - x_{m_k}\vert + \Vert g - y\Vert_q \Vert x \Vert_p \\ & \leq \Vert y - g \Vert_q \Vert x^n \Vert_p + n \max_{\{m_1, \ldots, m_k\}} \vert x^n_{m_k} - x_{m_k}\vert + \Vert g - y\Vert_q \Vert x \Vert_p \\ & <\frac{\varepsilon}{3 M} \Vert x^n \Vert_p + n\frac{\varepsilon}{3n} + \frac{\varepsilon}{3 \Vert x \Vert_p} \Vert x \Vert_p\\ & < \varepsilon \end{align}$$

The essence of this argument is that the condition $\forall m \in \mathbb{N}: \vert x^n_m - x_m \vert$ is precisely weak convergence for a dense set of functionals. Then it is only needed that $(x^n)_{n \in \mathbb{N}}$ has a uniform upper bound, which is crucial in bounding the term $\vert y(x^n) - g(x^n) \vert$.