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Let $X$ be a Banach space (it is in fact an $L^p$ space) and let $T:X \to X$ be a linear continuous operator (which is not injective and not surjective). I am trying to figure out if the following is true:

For every $g \in Im(T)$ there exist $f \in X$ such that $Tf=g$ and $\lvert \lvert f \lvert \lvert \leq C \lvert \lvert g \lvert \lvert$ for some $C>0$ independent of $g$.

I know that if $T$ was bijective then this would be a consequence of the Open Mapping Theorem (The inverse is continuous). But I cannot see any way to prove my claim from this theorem.

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This is not true in general. Let $X = \ell^2$, and $T \in L(\ell^2)$ given by $$ Tx = \left(\frac{x_n}{n+1}\right)_n $$ Then $T$ is one-to-one, but for $g = e_n \in \def\ran{\operatorname{im}}\ran T$ we have that the inverse image is $f = (n+1)e_n$, and $$ \|f\| = \|(n+1)e_n\| = (n+1)\|e_n\| \stackrel ?\le C \|e_n\| $$ Hence, such a $C$ cannot exist.

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  • $\begingroup$ Why does this not contradict the bounded inverse theorem? Isn't the inverse of a continuous linear bijective mapping supposed to be continuous? $\endgroup$ – Stef H Nov 16 '15 at 14:40
  • $\begingroup$ The inverse of a continuous linear bijective map between Banach spaces. Note that here $\ran T$ is not a Banach space. $\endgroup$ – martini Nov 16 '15 at 14:41
  • $\begingroup$ Oh, right, thank you. What happens if the image is closed, i.e., if $Im(T)$ is Banach? Is my claim true in this case? $\endgroup$ – Stef H Nov 16 '15 at 14:43
  • $\begingroup$ @StefH Such a $C$ exists if and only if $\operatorname{Im} T$ is closed. $\endgroup$ – Daniel Fischer Nov 16 '15 at 14:44
  • $\begingroup$ As @DanielFischer said, your claim is true then. Just look at the mapping $\hat T \colon X/\ker T \to \ran T$ induced by $T$ and apply the open mapping theorem $\endgroup$ – martini Nov 16 '15 at 14:45

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