# Continuous mapping with contractive n-th power

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.
• Why would the operator not be defined in all of $X$? – Martin Argerami Oct 4 at 20:01
• 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 5 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 9 at 3:27