# Banach's inverse operator theorem

1. Sep 6, 2014

### DavideGenoa

Dear friends, I have been trying in vain for a long time to understand the proof given in Kolmogorov and Fomin's of Banach's theorem of the inverse operator. At p. 230 it is said that $M_N$ is dense in $P_0$ because $M_n$ is dense in $P$.

I am only able to see the proof that $(P\cap M_n)-y_0 \subset P_0$ and that $(P\cap M_n)-y_0 \subset M_N$ there.
I obviously realise that $P_0=P-y_0$ and therefore $P_0\cap(M_n-y_0)=(P\cap M_n)-y_0 \subset M_N$, but I don't see why $P_0\subset\overline{M_N}$...
What I find most perplexing is that, in order to prove the density of $P_0$ in $M_N$, I would expect something like Let $x$ be such that $x\in P_0$... then $x\in \overline{M_N}$, while, there, we "start" from $z\in P\cap M_n$ such that $z-y_0\in P_0$, but I don't think that all $x\in P_0$ are such that $x+y_0\in P$... (further in the proof we look for a $\lambda$ such that $\alpha<\|\lambda y\|<\beta$, i.e. such that $\lambda y\in P_0$)

Has anyone a better understanding than mine? Thank you very much for any help!!!

2. Sep 7, 2014

### micromass

Staff Emeritus
Hint: $M_n\subseteq M_N$ so $M_N$ is also dense in $P$. And $P_0$ is an open subset of $P$.

3. Sep 7, 2014

### DavideGenoa

Thank you so much!!!
Forgive me: I don't know how to prove that...

4. Sep 7, 2014

### micromass

Staff Emeritus
Indeed, because it's not true. What I meant is that $P$ is an open subset of the sphere $S$. And $M_n$ (and thus $M_N$) are dense in there. Apologies for the inconvenience.

5. Sep 7, 2014

### DavideGenoa

Thank you very much: no problem for any mistyping!!! I see that $P$ is an open subset of the open sphere $S$, therefore $P\subset\overline{S}$, and $M_n$ is chosen such that $S\subset\overline{M_n}$, and $\overline{M_n} \subset\overline{M_N}$ since $M_n\subset M_N$, so I realise $M_N$ is dense in $P$ (because $P\subset\overline{S}\subset\overline{M_n}\subset\overline{M_N}$), but I don't see why $P_0\subset\overline{M_N}$...

6. Sep 7, 2014

### micromass

Staff Emeritus
$P_0$ is an open subset of the sphere $S$ too.

7. Sep 7, 2014

### DavideGenoa

Ehm... I cannot see that... Of course $y_0\in M_n\cap S$ and $P\subset S$, and therefore $P_0\subset S-y_0$, but I see nothing else relevant...

8. Sep 7, 2014

### micromass

Staff Emeritus
It's intuitive, no? We know that $P$ is (part of) a sphere inside the sphere $S$, but it's centered at $y_0$. Then we translate $P$ to be centered at the origin. I think it should be clear that this translation is still part of $S$.

9. Sep 7, 2014

### DavideGenoa

Thank you so much! Is $S$ is centred in $0$? In that case, yes, I see that $P_0\subset S$, because if it weren't so, then $\beta> r$ where $r$ is the radius of $S$, but, in that case, for any $\varepsilon>0$ we could find a $x\in P$ such that $\|x-y_0\|>\beta-\varepsilon>r-\varepsilon$ and, chosing $\varepsilon=\beta- r$, we would contradict $P\subset S$, I think. Though, I am not sure how we can chose a $S$ centred in $0$...

10. Sep 7, 2014

### micromass

Staff Emeritus
Haha, ok, $S$ is not necessarily centered in $0$. Missed that one.

OK, so $(4)$ basically proves that if $z\in P\cap M_n$, then $z-y_0\in P_0\cap M_N$.

So, to prove $M_N$is dense in $P_0$. Take an arbitrary $x\in P_0$. This is of the form $x = x^\prime - y_0$ with $x^\prime \in P$. Since $M_n$is dense in $P$, we can find a sequence $(x_n)_n\subseteq M_n\cap P$ such that $x_n\rightarrow x^\prime$. Then obviously by $(4)$, we have $(x_n - y_0)_n \subseteq P_0\cap M_N$ and $x_n - y_0\rightarrow x$

11. Sep 7, 2014

### DavideGenoa

If the book had used a handful of words more...
I deeply thank you... Now everything is clear.