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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.

pabodu
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1 Answers1

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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.

Martin Argerami
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  • 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. – pabodu Oct 04 '19 at 11:42
  • Why would the operator not be defined in all of $X$? – Martin Argerami Oct 04 '19 at 20:01
  • 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. – pabodu Oct 05 '19 at 12:56
  • 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. – Martin Argerami Oct 05 '19 at 14:39
  • 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 $T$. – pabodu Oct 09 '19 at 03:27
  • So to show something about an unbounded $T$, you assume it's bounded? Besides, it's not true that an intersection of closed nested sets in a Banach space is nonempty. Example: take $X=c_0$, and $V_n={x:\ x_1=\cdots=x_n=0}$. – Martin Argerami Oct 09 '19 at 03:43