Is this true? If $\varphi\in \mathcal{B}(E,F)$ and preserving density, then $\varphi$ is surjection, such that $E$ and $F$ are Banach space and $\mathcal{B}(E,F)$ is the set of all bounded linear operator from $E$ to $F$.
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Given $y\in F$, since $\varphi(E)$ is dense in $E$, there is a sequence $\{x_n\}\subset E$ such that $\lim_{n\to\infty}\varphi(x_n)=y$. This is not enough to conclude that there exists $x\in E$ such that $\varphi(x)=y$. In fact $\varphi$ may not be surjective.
Let $E=F=L^1[0,1]$ and $(\varphi f)(x)=\int_0^xf(t)\,dt$. We have $|(\varphi f)(x)|\le\|f\|_1$ and $\|\varphi f\|_1\le\|f\|_1$. $\varphi f$ is continuous for all $f\in E$, so that $\varphi$ is not surjective, but $\varphi(E)$ is dense in $f$.
Julián Aguirre
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Thank you very much, Professor Julian. I've edited my question, please guide me. – Alireza Feb 04 '15 at 11:18
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This is a different question. You should post a new question, and if my answer to the original question was helpful, upvote it and accept it. – Julián Aguirre Feb 04 '15 at 11:24
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Hi professor Julian. I posted a new question, can you help me, please? – Alireza Feb 05 '15 at 16:05