I'm proving the following:
Let $E,F$ be two Banach spaces; let $f$ be a function $f:E\to F\ $ linear, and such that: "for every sequence $(x_n)_{n\in\mathbb{N}}\subseteq E$ which converges to $x\in E$ and such that the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges to $y\in F$; we have $y=f(x)$." Then I need to prove that $f$ is continuous.
My approach: Step 1: Define $G=\{(x,f(x)):x\in E\}$ provide with the norm $||(x,f(x))||=||x||+||f(x)||$ we can prove that $(G,||\cdot||)$ is a Banach space.
Step 2: let $g:G\to E$ define by $g(x,f(x))=x$, so we can prove that: $g$ es linear, continuous, and 1-1 and onto, furthermore $||g||=\sup_{||(x,f(x))||=1}{||g(x,f(x))||}\leq 1$.
Step 3: It follows from the open mapping theorem that $g^{-1}$ exists and is continuous. But how can I prove that there is $C>1$ such that $C||g(y)||\geq ||y||$ for all $y\in G$? If I can prove this then the problem is done!!
Edit: thanks: I think i found was going on, $g^{-1}$ is continuous then exist $M>0$ such for all $x\in E$, $\|g^{-1}(x)\|\leq M\|x\| $ but then if $y=g^{-1}(x)$ we have $\|y\|\leq M\|g(y)\|$, for all $y$ in $G$. For other part, If $\|y\|=1=\|x\|+\|f(x)\|$ wich implice $\|x\|>1\Rightarrow M>1$ other case $x=0\ (\Rightarrow \ \Leftarrow)$. Finally $\|x\|+\|f(x)\|\leq M \|x\|$ with $M>1$, is the same to be continuous.
