The question could be very similar, but I am new in functional analysis.
Let us consider $l^2$ space (space of square summable sequences). Next, for each fixed $j = 1, 2, \dots$ let $X^{j}_{n} \in \mathbb{R}$ be a sequence, such that
$$
X^{j}_{n} \to c_{j} \in \mathbb{R},
$$
as $n \to \infty$.
Next, let us assume that $c\in l^2$, i.e. the vector comprised by the limits of the sequences is in $l^2$. Also, for any finite $n$ we have that $\{X^{j}_{n}\}_{j=1}^{\infty}$ is in $l^2$.
Summary: we have a sequence of infinite dimensional vectors $\{X^{j}_{n}\}_{j=1}^{\infty}$, such that each component $X^{j}_{n}$ converge to the number $c_{j}$ AND the vector $c=(c_{1}, c_{2}, \dots) \in l^{2}$ (point-wise convergence holds). Next, for all finite $n$ the vector is $\{X^{j}_{n}\}_{j=1}^{\infty}$ square summable.
Can we then claim that all elements of sequence $\{X^{j}_{n}\}_{j=1}^{\infty}$ are in $l^2$?
In another words, if the point-wise limit is in $l^{2}$ is it enough to claim the all sequence is in $l^{2}$? Note: I clearly understand, that convergence in $l^{2}$ does not follow from point-wise convergence.
UPDATE: Let us take upper right triangular infinite matrix with all elements ones. For fixed $n$ all columns up to $n$ are in $l^{2}$.
Can we say that ALL columns are in $l^{2}$?
Each raw converge though… This is how I came up to this question.
Best Answer
Note this is not an answer to the question as asked. As stated above, the question asks whether P implies P. I'm giving up on getting the OP the clarify what he or she actually meant to ask; this is an answer to what I suspect he or she meant to ask.
Let $X_n^j=\frac1{\sqrt{jn}}$ and $c_j=0$. Then $X_n\to c$ pointwise, $c\in\ell_2$ but $X_n\notin\ell_2$.
(Not an answer to the question as asked precisely because $X_n\in\ell_2$ is a hypothesis there...)