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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.