Let $X$ be a Banach space and $T$ is in the set of bounded linear operators from $X$ to $X$ and norm of $T$ is less than 1. Use contraction mapping principle show $I-T$ is an isomorphism.(I can show it is one to one, but not onto and continuous inverse)

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    $\begingroup$ Hint: $(1-x)^{-1} = 1 + x + x^2 + \cdots$ if $x \in (-1,1)$. Show that $$(I-T)(I + T + T^2 + \cdots) = I$$ if $\|T\| < 1$. The latter sum converging in the norm of ${\cal B}(X)$, of course. $\endgroup$ – Umberto P. Mar 8 '17 at 16:38
  • $\begingroup$ I still dont see the hint $\endgroup$ – dyyyyssss Mar 8 '17 at 16:53
  • $\begingroup$ I will be more explicit. Show that $$S = I + T + T^2 + \cdots$$ converges in the operator norm. Then show that $(I-T)S = I$ so that $I-T$ has a right-sided inverse. $\endgroup$ – Umberto P. Mar 8 '17 at 17:59
  • $\begingroup$ Actually $I-T$ is a linear homeomorphism by open mapping theorem. $\endgroup$ – Dbchatto67 Oct 7 '18 at 8:02

If $I -T$ is isomorphism if for all $y \in X$ there only one $x \in X$ such that $$x - Tx = y$$ or equivalently, $x$ is the only fixed point of map $f_y : X \to X$ given for $f_y(z) = Tz + y$. But, $$||f_y(z) - f_y(w)|| = ||Tz - Tw|| \leq ||T||||z - w||$$ that is, $f_y$ is a contraction and therefore has only one fixed point.

  • $\begingroup$ I dont think this helps $\endgroup$ – dyyyyssss Mar 8 '17 at 16:53
  • $\begingroup$ I'm only used contraction mapping principle show I−T is an isomorphism. The inverse you already showed. $\endgroup$ – A.D. Mar 8 '17 at 17:03
  • $\begingroup$ it makes sense, and how do you show it is onto? $\endgroup$ – dyyyyssss Mar 8 '17 at 17:09
  • $\begingroup$ It's because $\forall y \in X$ there is only one $x in X$ such that $f_y(x) = x$, but $f_y(x)= x$, this is $Tx + y = x$, if only if $y = x -Tx$ $\endgroup$ – A.D. Mar 8 '17 at 17:24
  • $\begingroup$ sorry, my coment was wrong, how do you show the inverse is continuous? $\endgroup$ – dyyyyssss Mar 8 '17 at 17:36

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