# Application of the open mapping theorem on sequences

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 $$x_0$$.

• What space is $A$ defined on? – user293794 Jan 12 at 21:33
• A is bounded linear operator from $X$ to $Y$. – Hana Jan 12 at 21:39

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.
• For every $n$ there exist some $C_n$. How to get unique $C$ for every $n$? – Hana Jan 14 at 16:08