Application of the open mapping theorem on sequences

Question

Let $X$ and $Y$ be Banach spaces and $A\in B(X,Y)$ surjective operator. I know that from open mapping theorem follow that there exist $C>0$ such that for every $y\in Y$ exist $x\in X$ such that $Ax=y$ and $||x||\leq C||y||$. Now I need to prove this for zero convergent sequence, i.e. exist $C>0$ such that for every sequence $\{y_n\}$ from $Y$ which converge to 0 exist sequence $\{x\}$ from $X$ which converge to 0 such that $Ax_n=y_n$ for every n and $||x_n||\leq C||y_n||$, and same statement when $\{y_n\}$ converge to $y_0$ and $\{x_n\}$ converge to $x_0$.

Answer 1

From the open mapping theorem, we know that for every $y_n$ in your sequence there exists $x_n$ such that $Ax_n=y_n$ and $||x_n||\leq C||y_n||$. If $y_n\rightarrow 0$ then so does $x_n$ from the inequality $||x_n||\leq C||y_n||$. Now if $y_n\rightarrow y_0$, instead consider the sequence $z_n:=y_n-y_0$, which goes to $0$, and reason as before.