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Let $A \colon X \to X$ be a bounded linear operator, where $X$ is a Banach space.

$(Q1)$ Is it true that if $A$ is injective then the image of $A$ is dense in $X$?

$(Q2)$ Is it true that $A^{-1} \colon \text{rgA} \to X$ is closed?

My attempts of solution:

$(A1)$ I have no idea.

$(A2)$ I know that if $A$ is bounded then it is closed, i.e. if $$x_n \to x,\ \text{and}\ Ax_n \to y,\ \Longrightarrow x \in D(A)( = X)\ \text{and}\ Ax = y.$$ Let $y_n \to y$ and $A^{-1}y_n \to w$. $(*)$

By definition, there exist $\{x_n\}$ such that $Ax_n = y_n$ substituting this into $(*)$ we get $$Ax_n \to y,\ \text{and}\ A^{-1}Ax_n = x_n \to w.$$

By closedness of $A$ this implies that $w \in X$ (obviously) and that $Aw = y$, i.e. $w = A^{-1}y$.

Do you have any idea for $(Q1)$? is $(A2)$ correct?

Thank you very much for your time and help!!!

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1 Answer 1

up vote 1 down vote accepted

(A2) is correct. Note that it assumes that $A$ be injective. And that $X$ be a normed vector space suffices.

For (Q1), a natural family of counterexamples is given by non surjective isometries on a Banach space, since their ranges are closed proper subspaces. For instance, the unilateral shift operator $S:\ell^2(\mathbb{N})\longrightarrow \ell^2(\mathbb{N})$ $$ S:(x_0,x_1,x_2,\ldots)\longmapsto (0,x_0,x_1,x_2,\ldots). $$

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thank you! I'll have to keep in mind that the shift is always the answer! :D –  user01123581321345589144... May 14 '13 at 18:06
1  
@user01123581321345589144... You're welcome. Yes, the shift does provide a lot of examples and counterexamples. –  1015 May 14 '13 at 18:12

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