
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 $\displaystyle X$ by $\displaystyle \cdot_X$, and balls by $\displaystyle B_X(x,r)$. The interior of a set $\displaystyle A$ is $\displaystyle A^o$ and its closure $\displaystyle \overline{A}$.
Baire's Lemma. IF $\displaystyle X$ is a complete metric space, and $\displaystyle (X_n)$ is a sequence of closed subsets of $\displaystyle X$ with $\displaystyle X=\cup_n X_n$, then there exists an index $\displaystyle n_0$ with $\displaystyle X^o_{n_0}\neq \emptyset$.
It's philosophy is simple: if the result didn't hold, then we would obtain $\displaystyle \cap_n X^c_{n}=\emptyset$, violating the nested sets property, and so $\displaystyle X$ would not be complete.
We can now prove
Open Mapping Theorem. Let $\displaystyle T:X\rightarrow Y$ be a linear, continuous, injective and surjective operator between two Banach spaces. Then $\displaystyle T$ maps open sets in $\displaystyle X$ to open sets in $\displaystyle Y$.
Proof.
We need to show, that if $\displaystyle A$ is open in $\displaystyle X$, then $\displaystyle T(A)$ is open in $\displaystyle Y$. Crumble this into pieces:
Claim 1. We need only consider the case where $\displaystyle A$ is an open ball.
Truly, suppose the result holds in this case. if $\displaystyle A$ is any open set, there exists a family of open balls $\displaystyle (B_X(a,r_a))_{a\in A}$ such that $\displaystyle A=\cup_{a\in A} B_X(a,r_a)$, as open balls constitute a basis for the topology of $\displaystyle X$. Then, $\displaystyle T(A)=\cup_a T(B_X(a,r_a))$, and so $\displaystyle T(A)$ is open in $\displaystyle Y$ 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 $\displaystyle x_0\in X$ we shall prove that the ball
$\displaystyle B(x_0,1)=x_0+B_X(0,1)=\{x_0+v: \ v_X<1\}$ is mapped into an open set containing $\displaystyle T(x_0)$. Linearity gives
$\displaystyle T(x_0+B_X(x_0,1))=T(x_0)+T(B_X(0,1))$
and by the hypothesis, $\displaystyle T(B_X(0,1))$ is open. So $\displaystyle T(x_0)+T(B_X(0,1))$ is an open set containing $\displaystyle T(x_0)$.
Claim 3. There exists $\displaystyle r>0$ with $\displaystyle B_Y(0,2r)\subset \overline{T(B_X(0,1))}$
Time for Baire's Lemma. Let $\displaystyle X_n=n\overline{T(B_X(0,1))}, \ n \in {\bf N}.$ Since $\displaystyle T$ is surjective, we have $\displaystyle X=\cup_n X_n$. So, there exists $\displaystyle n_0$ with $\displaystyle X_{n_0}^o\neq \emptyset.$ Especially, $\displaystyle \overline{[T(B_X(0,1))]}^o\neq \emptyset$.
Since this is open in $\displaystyle Y$, we can find a ball $\displaystyle B_Y(y,4r)\subset\overline{T(B_X(0,1))}$. Note that linearity guarantees $\displaystyle y,y\in\overline{T(B_X(0,1))}$. This means
$\displaystyle y+B_Y(y,4r)=B_Y(0,4r)\subset \overline{T(B_X(0,1))}+\overline{T(B_X(0,1))},$
but also
$\displaystyle \overline{T(B_X(0,1))}+\overline{T(B_X(0,1))}=2 \overline{T(B_X(0,1))}$
as this set is convex. So $\displaystyle B_Y(0,4r)\subset 2\overline{T(B_X(0,1))}$.
That is, if $\displaystyle w\in B_Y(0,4r)$, then $\displaystyle w=2\lim T(w_n)$ for some sequence $\displaystyle (w_n)\subset B_X(0,1)$. Then $\displaystyle w_Y=\limT(w_n)_Y\leq 2r$, so $\displaystyle B_Y(0,2r)\subset \overline{T(B_X(0,1))}$.
Claim 4. It holds that $\displaystyle B_Y(0,r)\subset T(B_X(0,1))$
With this, the proof will be complete.
Choose $\displaystyle y\in B_Y(0,r).$ We have by Claim 3:
There exists $\displaystyle x_1\in X$ with
$\displaystyle x_1_X< 2^{1}, \ yT(x_1)_Y<2^{1}r$
Use Claim 3 again, to pick $\displaystyle x_2\in X$ with
$\displaystyle x_1_X< 2^{2}, \ yT(x_1+x_2)_Y< 2^{2}r$
and continue inductively, to construct a sequence $\displaystyle x_n$ such that
$\displaystyle x_n_X < 2^{n}, \ yT(x_1+\ldots+x_n)_Y< 2^{n}r \ \ (1)$.
Let
$\displaystyle x^*_n=\sum_{k=1}^{n} x_k.$ Then, simple calculus gives us that $\displaystyle (x_n^*)$ is a Cauchy sequence, and must therefore converge to an element
$\displaystyle x\in X$ such that
$\displaystyle x_X<\sumx_n_X\leq \sum2^{n}=1$ (by the first inequality of (1))
and $\displaystyle yT(x_n^*)_Y\leq 2^{n}r\Rightarrow \lim T(x_n^*)=T(x)=y$ (by the second inequality of (1)). Thus $\displaystyle y\in T(B(0,1))$.
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 $\displaystyle T^{1}$ is also continuous.
Proof. By Claim 4, every $\displaystyle x\in X$ with $\displaystyle T(x)_Y< r$ satisfies $\displaystyle x_X<1$. So $\displaystyle x_X\leq r^{1}Tx_Y$, which means that the inverse operator $\displaystyle T^{1}$ is continuous.
Porisma 2. Consider a linear space $\displaystyle S$, endowed with two norms $\displaystyle \cdot_1, \ \cdot_2$, such that $\displaystyle S$ is a banach space for either norms. If there exists $\displaystyle c>0$ with $\displaystyle x_2\leq cx_1 \ \forall x\in X$, then there exists $\displaystyle c^*>0$ with $\displaystyle x_1\leq c^*x_2 \ \forall x\in X$ .
Proof. Apply Porisma 1 for $\displaystyle X=(S,\cdot_1)$, $\displaystyle Y=(S,\cdot_2)$ and $\displaystyle T$ the identity mapping.
(Note. $\displaystyle c^*$ 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 $\displaystyle T:X\rightarrow Y$ be a linear operator between two Banach spaces. If its graph,
$\displaystyle G(T)=\{(x,Tx): \ x\in X\}$
is a closed subset in the space $\displaystyle X\times Y$, then $\displaystyle T$ is continuous.
Proof.
Consider $\displaystyle X$ equipped with the norms
$\displaystyle x_1=x_X+Tx_Y, \ x_2=x_X.$
Since $\displaystyle G(T)$ is closed, $\displaystyle X$ is a Banach space for either norm. We observe that $\displaystyle x_2\leq x_1$; thus Porisma 2 implies
$\displaystyle x_1=x_X+Tx_Y\leq cx_2=cx_X$,
that is
$\displaystyle Tx_Y\leq (c1)x_X \ \forall x\in X$,
so $\displaystyle T$ 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