Let E be a Banach space.

Let $L(E)$ be the set of linear continuous operators from E to E.

Let A be the set of inversible elements of $L(E)$.

Prove that A is open in $L(E)$ and that $T→ T^{−1}$ is smooth in $L(E)$.

What I did:

I proved that if $||T|| < 1$ then $Id − T ∈ A$

I proved also that if $T ∈ A$ and $S ∈ L(E)$ then $T + S ∈ A$.

Now I need to prove that A is open so I need to find a ball of center $T$ such that all operators S belonging to this ball are in A.

For the smoothness of the inverse operator in A, I have no clue on how to prove it.

Thank you for any help or hints

  • 1
    $\begingroup$ You didn't prove that if $T\in A$ and $S\in L(E)$, then $T+S\in A$. Not for an arbitrary $S$. This will only hold for $S$ with small norm. You can prove that the latter holds by doing $T+S=T(I+T^{-1}S)$. Therefore, for $\|T^{-1}S\|<1$, you will have that $T+S$ is invertible, since $I+T^{-1}S$ is invertible. An explicit bound on the norm of $S$ could be $\|S\|<\|T^{-1}\|^{-1}$. $\endgroup$ – logarithm Apr 30 at 18:22
  • $\begingroup$ For smoothness you can use the same idea. The elements of $A$ near $T$ are parametrized by $T+S$ and their inverses by the convergent power series $(I+T^{-1}S)^{-1}T=\sum_{n=0}^{\infty}(-1)^n(T^{-1}S)^nT$. $\endgroup$ – logarithm Apr 30 at 18:28
  • $\begingroup$ Thank you @logarithm. I indeed need that notion of S close enough to T. But how to express clearly that the bound $||ST^{-1}||^{-1}$ means $||T-S ||< \delta$ $\endgroup$ – PerelMan Apr 30 at 19:31
  • 1
    $\begingroup$ It is not $S$ close to $T$, it is $T+S$ close to $T$, or what is equivalent $S$ small. $\endgroup$ – logarithm Apr 30 at 19:36

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