Is this true? If ‎$‎\varphi\in ‎\mathcal{B}(E,F)‎$‎ and preserving density, then $‎\varphi‎$‎ is surjection. Why?

Question

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 $F$.

Answer 1

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 $f$.