Existence of bounded sequences for image of bounded linear operators Let $X$ and $Y$ be Banach spaces and $A:X\to Y$ be a bounded linear operator. Assume that $y\in \overline {A(X)}$. Can we always choose a  bounded sequence $(x_n)\subset X$ such that $\displaystyle \lim_{n\to \infty} Ax_n=y$?
 A: Consider the operator $A: \ell_2\rightarrow \ell_2$ for which $A((x_n)_{n\in \mathbb{N}})=(\tfrac{x_n}{n})_{n\in \mathbb{N}}$ and take $\tilde{y}=(\tfrac{1}{n})_{n\in\mathbb{N}}$. This is an element of $\ell_2$ which does not belong in the image of $A$, however it belongs in $\overline{A(\ell_2)}$  as $A(\ell_2)$ is dense in $\ell_2$. Our goal is to show that every sequence $(\tilde{x}_n)_{n\in \mathbb{N}}$ in $\ell_2$ such that $A\tilde{x}_n\rightarrow \tilde{y}$, is unbounded. 
Lets write each $\tilde{x}_n$ as $\tilde{x}_n=(x_k^n)_{k\in\mathbb{N}}$.  The idea is that since $\|A\tilde{x}_n-\tilde{y}\|_2^2=\sum_{k=1}^\infty \frac{(x^n_k-1)^2}{k^2}\underset{n\rightarrow \infty}{\longrightarrow} 0$, a lot of the numbers $x_k^n$ will have to be "near" $1$. But these $x_k^n$ will inflate  the corresponding norm $\|\tilde{x}_n\|_2= \left(\sum_{k=1}^\infty (x_k^n)^2\right)^{1/2}$. 
Let $N_0\in \mathbb{N}$. For $\varepsilon = \sum_{k=N_0}^\infty \frac{1}{4k^2}$, there exists an $n_0$ such that 
$$ \sum_{k=1}^\infty \left(\frac{x_k^n-1}{k}\right)^2<\varepsilon=\sum_{k=N_0}^\infty \frac{1}{4k^2}, \ \ \forall n\geq n_0.$$
We consider the sets $A_n=\{k\in \mathbb{N}: x_k^n\geq \tfrac{1}{2}\}$. Then the cardinality of each $A_n$ is at least $\#A_n\geq N_0$. If not, then 
$$\sum_{k=1}^\infty \left(\frac{x_k^n-1}{k}\right)^2 \geq \sum_{k\notin A_n} \left(\frac{x_k^n-1}{k}\right)^2 \geq \sum_{k\notin A_n} \frac{ 1}{4k^2}\geq \sum_{k=N_0}^\infty \frac{ 1}{4k^2}=\varepsilon, $$
a contradiction.
So, for every $n\geq n_0$, $\|\tilde{x}_n\|_2\geq \left(\sum_{k\in A_{n}} \frac{1}{4}\right)^{1/2}\geq \frac{\sqrt{N_0}}{2}$. Since $N_0$ was arbitrary, $(\tilde{x}_n)_{n\in\mathbb{N}}$ is unbounded.
