Existence of surjective linear operator

[DavideGenoa]

Dear friends, I read that, if $A$ is a bounded linear operator 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!

 

[Fredrik]

In other words, you want to prove that there's an open ball around A such that every operator in it is surjective onto $E_1$.

I too would interpret their choice of words as saying that $A(E)=E_1$ (i.e. A is surjective onto $E_1$), and probably also that A is injective. I don't think we would say that, for example, the operator on $\mathbb R^3$ that projects onto the z axis "transforms" $\mathbb R^3$ to $\mathbb R$.

I think I found a solution that works for Hilbert spaces, based on the theorem that says that when $T$ is a bounded linear operator such that $\|1-T\|<1$, then $T$ is invertible and the inverse is given by the geometric series $T^{-1}=\sum_{k=1}^\infty T^k$. Unfortunately I don't know much about Banach spaces beyond their definition.

 

[DavideGenoa]

I too would interpret their choice of words as saying that $A(E)=E_1$ (i.e. A is surjective onto $E_1$), and probably also that A is injective. I don't think we would say that, for example, the operator on $\mathbb R^3$ that projects onto the z axis "transforms" $\mathbb R^3$ to $\mathbb R$.

In other parts of the book, an Italian language translation of A.N. Kolmogorov and S.V. Fomin's Элементы теории функций и функционального анализа, the verb to transform is used even for non-surjective operators, but, here, I wouldn't consider the lemma to be proven as true if $A$ weren't surjective, since $\forall\alpha>0\quad \|A-A\|<\alpha$. In the case $A$ were bijective, the lemma would be the same as "books.google.com/books?id=cbbCAgAAQBAJ&pg=PA231#v=onepage&q&f=false" , which precedes it by two pages in my book, but that would be quite strange...

I think I found a solution that works for Hilbert spaces, based on the theorem that says that when $T$ is a bounded linear operator such that $\|1-T\|<1$, then $T$ is invertible and the inverse is given by the geometric series $T^{-1}=\sum_{k=1}^\infty T^k$. Unfortunately I don't know much about Banach spaces beyond their definition.

If $T:E\to E$ is a bounded linear operator mapping a Banach space into itself, an identical theorem also holds for any Banach space, cfr. A.N. Kolmogorov, S.V. Fomin, Introductory real analysis, p. 232. How could it be used to prove the given lemma?
Thank you very much, Fredrik!

 

[Fredrik]

If A is not injective, I don't know. I interpreted the question as being about an invertible A. This is what I did:

I started by trying to rewrite A-B in some other way, e.g. $A-B=A(I-A^{-1}B)$. This didn't immediately solve the problem, but I realized that if $A^{-1}B$ is invertible, then $I=(A^{-1}B)^{-1}A^{-1}B$, and this ensures that B is invertible with inverse $(A^{-1}B)^{-1}A^{-1}$. So we just need to show that $A^{-1}B$ is invertible.

If $\|A-B\|<\|A\|$ we have
$$\|I-A^{-1}B\|=\|A^{-1}(A-B)\|\leq\|A^{-1}\|\|A-B\|=\frac{1}{\|A\|}\|A-B\|<1.$$ So the choice $\alpha=\|A\|$ gets the job done.

Edit: I made a mistake before, and wrote $\alpha=1/\|A\|$ instead of $\alpha=\|A\|$. I have edited that above.

 

[DavideGenoa]

Thank you so much!