Assume we have a normed space $X$ such that $A\in B(X)$ has the property that for every $\{x_n\}_{n=1}^{\infty}$ which is bounded in $x$, there is a subsequence such that $\{Ax_{n_k}\}_{k=1}^{\infty}$ is convergent in $X$. Denote by $Y$ the banach space which is the completion of $X$, and denote by $\tilde{A}$ the extension of $A$ to $Y$.
Is it necessarily true that $\tilde{A}$ is compact?
I think it is, but I'm not quite sure if my arguments are sufficient. I think that since $A$ is compact in $X$, we get that $A(X)_1$ is precompact, and $A(X)_1\subset\tilde{A}(Y)_1\subset\overline{A(X)_1}$ so it is also precompact. Is this statement even true?
Take $y_n$ a bounded sequence in $Y$. $X$ is embedded isometrically into a dense subset of $Y$, therefore or all $ \epsilon >0$ there exists a sequence $x_n$ in $X$ such that for all $n \in \mathbb N$ $$ || y_n - x_n|| < \epsilon$$ In particular also the sequence $x_n$ is bounded in $X$. Now take $x_{n_k}$ a subsequence of $x_n$ such that $Ax_{n_k}$ converges to an element $z \in X$. It follows that also $\tilde{A}y_{n_k}$ converges to $z$, in fact taking $N \in \mathbb{N}$ such that for $k >N$ we have $|| Ax_{n_k} - z||<\epsilon$, then for all $k>N$ we have $$ || \tilde{A}y_{n_k} - z|| \leq ||\tilde{A}y_{n_k} - Ax_{n_k}|| + || Ax_{n_k} - z|| \leq ||\tilde{A}|| || y_{n_k} - x_{n_k}|| + || Ax_{n_k} - z|| < ||A||\epsilon + \epsilon $$