
open mapping theorem
Hi
this is rosana
I need your help as soon as you can
I need the explaination to proof of open mapping theorem( letT be a continuous linear operator on an FspaceX onto an Fspace Y. then T maps every open set of X onto an open set of Y ) in Functional analysis for kosaku yoside , I am so glad for doing that for me
I try to understand the proof from this book but i can not
or if you have any proof to this theorem in simplist way from any where, give it to me please
thanks alot for any responding :)

I don't think you can find a proof that is "too simple". :(
Any proof would have to use an equivalent to Baire's Lemma  besides that, it's relatively simple manipulation. Are your trouble concerned with Baire's Lemma?

thanks a lot for responding
I really get in trouble to understand this proof so , if you can give to me simple manipulation that you Know ,I am so greatfull
because i am not understand the mean of Baire's Lemma also
I need your help as soon as you can :)

Let's see...
The following proofs follow the excellence of Royden's "Real Analysis", with a little delicate touches by Brezis' "Functional Analysis".
(...and loads of bulkwork from Rebesques' infamous "Indifferential Geometry").
Denote norms in space by , and balls by . The interior of a set is and its closure .
Baire's Lemma. IF is a complete metric space, and is a sequence of closed subsets of with , then there exists an index with .
It's philosophy is simple: if the result didn't hold, then we would obtain , violating the nested sets property, and so would not be complete.
We can now prove
Open Mapping Theorem. Let be a linear, continuous, injective and surjective operator between two Banach spaces. Then maps open sets in to open sets in .
Proof.
We need to show, that if is open in , then is open in . Crumble this into pieces:
Claim 1. We need only consider the case where is an open ball.
Truly, suppose the result holds in this case. if is any open set, there exists a family of open balls such that , as open balls constitute a basis for the topology of . Then, , and so is open in as a union of open sets.
Claim 2. We need only prove this for the unit ball, centered at the origin.
Again, suppose the result holds in this case. Then, for any we shall prove that the ball
is mapped into an open set containing . Linearity gives
and by the hypothesis, is open. So is an open set containing .
Claim 3. There exists with
Time for Baire's Lemma. Let Since is surjective, we have . So, there exists with Especially, .
Since this is open in , we can find a ball . Note that linearity guarantees . This means
but also
as this set is convex. So .
That is, if , then for some sequence . Then , so .
Claim 4. It holds that
With this, the proof will be complete.
Choose We have by Claim 3:
There exists with
Use Claim 3 again, to pick with
and continue inductively, to construct a sequence such that
.
Let
Then, simple calculus gives us that is a Cauchy sequence, and must therefore converge to an element
such that
(by the first inequality of (1))
and (by the second inequality of (1)). Thus .
Onwards to the Closed Graph Theorem. We shall need the following two simple porismata:
Porisma 1. If the conditions of the Open Mapping Theorem hold, then is also continuous.
Proof. By Claim 4, every with satisfies . So , which means that the inverse operator is continuous.
Porisma 2. Consider a linear space , endowed with two norms , such that is a banach space for either norms. If there exists with , then there exists with .
Proof. Apply Porisma 1 for , and the identity mapping.
(Note. can be made larger at will.)
( This Porisma actually establishes that the two norms are equivalent. So, it is sufficient to bound one norm in terms of the other, to obtain equivalence.)
And finally, we can prove
Closed Range Theorem. Let be a linear operator between two Banach spaces. If its graph,
is a closed subset in the space , then is continuous.
Proof.
Consider equipped with the norms
Since is closed, is a Banach space for either norm. We observe that ; thus Porisma 2 implies
,
that is
,
so is continuous.

miss_lolitta is here
Rebesques thanks soooooosooooooo :p much
rozana good luck ;)

Rebesques thanks sooooooooooooooooooooooooo much for your caring
and helping me in this difficult theorem :D
hiiiiiiiiiiiiiiii miss lolita
thanks for inter this amazing theorem ;)

Are the two of you related in any way? The math bulb in my head is lit with suspicions of a pattern here.
Anyway... Nice course this must be.

Rebesques: you understand it on the fly..
that's right,Rosana is my friend
The math bulb in our heads has a good light to guess truths :D
take care!! ;)

hiiiiiiiiiiiiiiiiiiiiiii
I believe that every one who enter this field has a great mind like you Repesque . ;)
missloltta my best friend :p
Good Luck