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Given $N$-th power of mapping, defined on all Banach space, is a contraction, is the mapping continuous?\

Solution. Let for $N=2$, $T^2$ be contractive. Then $T^2$ is continuous. For sequence $x_n\to x$ we consider $Tx_n$ and $Tx$ and assume $Tx_n\not\to Tx$. Since $T^2x_n\to T^2x$, then we need to show that inverse $T^{-1}$ is continuous, then $T^{-1}T^2x=Tx$. But I could not show or disprove that $T^{-1}$ is continuous.

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No, not even if $T$ is linear. Let $X$ be any infinite-dimensional Banach space. Let $f$ be an unbounded linear functional. Fix $e_1\in X$ nonzero; then the subspace $\mathbb C\,e_1$ is complemented, call the complement $Y$. That is, $X=\mathbb C\,e_1+Y$ as a direct sum. Define $T$ by $$ T(\lambda e_1+y)=f(y) e_1 . $$ Then $T$ is linear, unbounded, and $$ T^2(\lambda_1 e_1+y)=T(f(y) e_1 +0) =0. $$ So $T^2=0$, a contraction.

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  • $\begingroup$ This is a nice example. However, your operator isn't defined on all Banach space. My question is about mappings defined on all Banach space. So, I corrected the question. Thank you. $\endgroup$ – pabodu Oct 4 at 11:42
  • $\begingroup$ Why would the operator not be defined in all of $X$? $\endgroup$ – Martin Argerami Oct 4 at 20:01
  • $\begingroup$ I don't know any sample of an unbounded operator defined on the entire Banach space. Usually, people figure out the domain of the unbounded operator, which is dense in the Banach space. $\endgroup$ – pabodu Oct 5 at 12:56
  • $\begingroup$ Unless you want to not use the Axiom of Choice (and then you don't have Hahn-Banach, and most of the classic Functional Analysis theorems) constructing an unbounded operator on all of $X$ is very easy. Note that the only hypothesis to construct an unbounded linear functional there is that every vector space has a basis. $\endgroup$ – Martin Argerami Oct 5 at 14:39
  • $\begingroup$ I think you are not right. Unbounded operators in Banach space have a proper subset as the domain. Really, Let $V_1=\{x:\ ||Tx||\geq1\}$, $V_2=\{x:\ ||Tx||\geq2\}$, $\ldots$, $V_n=\{x:\ ||Tx||\geq n\}$, $\ldots$,\\ Then $V_1\supset V_2\supset\ldots\supset V_n\ldots$.\\ The intersection $\bigcap_n V_n$ of closed nested sets in Banach space has nonempty intersection. Operator $T$ is not defined on that intersection. Sets $V_n$ are closed thanks to the continuity of operator $T$. $\endgroup$ – pabodu Oct 9 at 3:27

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