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Joonas Ilmavirta
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Davide Giraudo
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Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $dom(A)=range(A)$$\mathrm{dom}(A)=\mathrm{range}(A)$, $dom(A)$$\mathrm{dom}(A)$ dense in $B$.

Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity?

Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $dom(A)=range(A)$, $dom(A)$ dense in $B$.

Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity?

Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $\mathrm{dom}(A)=\mathrm{range}(A)$, $\mathrm{dom}(A)$ dense in $B$.

Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity?

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CPJ
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Can the boundedness of $A^2$ imply the boundedness of $A$?

Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $dom(A)=range(A)$, $dom(A)$ dense in $B$.

Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity?