Dense Domain: Preimage

Question

Given Banach spaces $X$ and $Y$.

Regard a bounded operator: $$A\in\mathcal{B}(X,Y)\implies A\in\mathcal{C}(X,Y)$$

Then for dense sets: $$W\leq Y:\quad \overline{W}=Y\implies\overline{A^{-1}W}=X$$

How can I see this?

Answer 1

Let $Af = \int_{0}^{x}f(t)dt$ in $L^{2}[0,1]$. Then $A : L^{2}\rightarrow L^{2}$ is bounded. Let $W$ consist of all continuously differentiable $g \in L^{2}[0,1]$ for which $g(0)=g(1)=0$. $W$ is dense in $L^{2}[0,1]$ because $\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$ is an orthogonal basis of $L^{2}[0,1]$. However, $A^{-1}W$ is not dense because $f \in A^{-1}W$ implies $(f,1)=0$.