Dear friends, I read that, if ##A## is a bounded linear operator(adsbygoogle = window.adsbygoogle || []).push({}); transforming-I think that such a terminology implies that ##A## is surjective because if ##B=A## and ##A## weren't surjective, that would be a counterexample to the theorem; please correct me if I'm wrong- a Banach space ##E## into a Banach space ##E_1##, there is a constant ##\alpha>0## such that, if ##B\in\mathscr{L}(E,E_1)## is a continuous linear operator defined in ##E## and ##\|A-B\|<\alpha##, then ##B## is surjective.

I thought I could use the Banach contraction principle, but I get nothing...

##\infty## thanks for any help!!!

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Existence of surjective linear operator

Loading...

Similar Threads - Existence surjective linear | Date |
---|---|

I Existence of Partial Derivatives and Continuity ... | Feb 17, 2018 |

I Proving that square root of 2 exists | Sep 24, 2017 |

I Injective & Surjective Proofs | Mar 8, 2017 |

How many topologies exist on 4 points? Any nomenclature? | Jul 11, 2015 |

How do I prove the existence of this norm? | Jun 1, 2015 |

**Physics Forums - The Fusion of Science and Community**