3

This is Problem 26, Chapter 4 in Rudin's Functional Analysis

$X, Y$ are Banach Spaces and $\mathcal{B}(X, Y)$ denotes the collection of all continuous linear maps from $X$ to $Y$.

Assume $T \in \mathcal{B}(X, Y)$ and $T(X) = Y$. Show that there is a $\delta > 0 $ such that $S(X) = Y$ for all $S \in \mathcal{B}(X, Y)$ with $|| S - T || < \delta$

By Open Mapping theorem this is equivalent to $T(B_X) \supset rB_Y$ for some $r > 0$ where $B$s denote the open unit balls.

This result is of course true in the finite dimensional case.

I have proved a similar result: Collection of Compact operators(these are never surjective) form a closed set but I have not made any progress on this one.

I am looking for hints if possible.

Subham Jaiswal
  • 3,593
  • Are you familiar with Neumann series? It states that if $R:X \to X$ is a bounded linear operator with $||R|| < 1$ then $\mathrm{Id}-R$ is invertible and the inverse equals the convergent geometric series $\sum_k R^k$. Try to reduce your setting to this case by applying $T^{-1}$ in a clever way, which is bounded by the open mapping theorem. – Hidde Apr 29 '23 at 06:55
  • @Hidde I am able to see that the invertible elements form an open subset. But the given $T$ is not invertible. We do have a open nbhd around $T'$ in $(\frac{X}{kerT})^$ but that doesn't give an open set in X. I think I am not seeing what you have jn mind – Subham Jaiswal Apr 29 '23 at 07:45
  • A proof can be performed by mimicking the proof of the open mapping theorem. – Ryszard Szwarc Apr 29 '23 at 08:10
  • 1
    @SubhamJaiswal Good point, I mistakenly assumed $T$ was injective. Still a similar argument works. If $y_0 \in Y$ then we can define an iterative sequence $y_{n+1}=y_n-Sx_n$ with $x_n \in X$ chosen such that $Tx_n=y_n$ and $||x_n||\leq||y_n||/r$. Then show that $y=\sum_n y_n$ and $x=\sum_n x_n$ converges and $Sx=y_0$ if $||T-S||$ is sufficiently small. – Hidde Apr 29 '23 at 08:37
  • @Hidde wow! That's a beautiful idea. – Subham Jaiswal Apr 29 '23 at 09:38
  • Have a look here: https://math.stackexchange.com/questions/17087/why-is-the-space-of-surjective-operators-open – Gerd Apr 30 '23 at 06:26

0 Answers0