0
$\begingroup$

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 $x_0$.

$\endgroup$
  • $\begingroup$ What space is $A$ defined on? $\endgroup$ – user293794 Jan 12 at 21:33
  • $\begingroup$ A is bounded linear operator from $X$ to $Y$. $\endgroup$ – Hana Jan 12 at 21:39
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.