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!!!

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# Existence of surjective linear operator

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