open mapping theorem

**rosana** · April 15th 2006, 01:21 AM

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 F-spaceX onto an F-space 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

**Rebesques** · April 17th 2006, 03:39 PM

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?

**rosana** · April 18th 2006, 10:28 PM

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

**Rebesques** · April 19th 2006, 12:51 PM

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 $X$ by $||\cdot||_X$ , and balls by $B_X(x,r)$ . The interior of a set $A$ is $A^o$ and its closure $\overline{A}$ .

Baire's Lemma. IF $X$ is a complete metric space, and $(X_n)$ is a sequence of closed subsets of $X$ with $X=\cup_n X_n$ , then there exists an index $n_0$ with $X^o_{n_0}\neq \emptyset$ .

It's philosophy is simple: if the result didn't hold, then we would obtain $\cap_n X^c_{n}=\emptyset$ , violating the nested sets property, and so $X$ would not be complete.

We can now prove

Open Mapping Theorem. Let $T:X\rightarrow Y$ be a linear, continuous, injective and surjective operator between two Banach spaces. Then $T$ maps open sets in $X$ to open sets in $Y$ .

Proof.
We need to show, that if $A$ is open in $X$ , then $T(A)$ is open in $Y$ . Crumble this into pieces:

Claim 1. We need only consider the case where $A$ is an open ball.

Truly, suppose the result holds in this case. if $A$ is any open set, there exists a family of open balls $(B_X(a,r_a))_{a\in A}$ such that $A=\cup_{a\in A} B_X(a,r_a)$ , as open balls constitute a basis for the topology of $X$ . Then, $T(A)=\cup_a T(B_X(a,r_a))$ , and so $T(A)$ is open in $Y$ 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 $x_0\in X$ we shall prove that the ball
$B(x_0,1)=x_0+B_X(0,1)=\{x_0+v: \ ||v||_X<1\}$ is mapped into an open set containing $T(x_0)$ . Linearity gives
$T(x_0+B_X(x_0,1))=T(x_0)+T(B_X(0,1))$
and by the hypothesis, $T(B_X(0,1))$ is open. So $T(x_0)+T(B_X(0,1))$ is an open set containing $T(x_0)$ .

Claim 3. There exists $r>0$ with $B_Y(0,2r)\subset \overline{T(B_X(0,1))}$

Time for Baire's Lemma. Let $X_n=n\overline{T(B_X(0,1))}, \ n \in {\bf N}.$ Since $T$ is surjective, we have $X=\cup_n X_n$ . So, there exists $n_0$ with $X_{n_0}^o\neq \emptyset.$ Especially, $\overline{[T(B_X(0,1))]}^o\neq \emptyset$ .

Since this is open in $Y$ , we can find a ball $B_Y(y,4r)\subset\overline{T(B_X(0,1))}$ . Note that linearity guarantees $y,-y\in\overline{T(B_X(0,1))}$ . This means

$-y+B_Y(y,4r)=B_Y(0,4r)\subset \overline{T(B_X(0,1))}+\overline{T(B_X(0,1))},$

but also
$\overline{T(B_X(0,1))}+\overline{T(B_X(0,1))}=2 \overline{T(B_X(0,1))}$

as this set is convex. So $B_Y(0,4r)\subset 2\overline{T(B_X(0,1))}$ .
That is, if $w\in B_Y(0,4r)$ , then $w=2\lim T(w_n)$ for some sequence $(w_n)\subset B_X(0,1)$ . Then $||w||_Y=\lim||T(w_n)||_Y\leq 2r$ , so $B_Y(0,2r)\subset \overline{T(B_X(0,1))}$ .

Claim 4. It holds that $B_Y(0,r)\subset T(B_X(0,1))$

With this, the proof will be complete.

Choose $y\in B_Y(0,r).$ We have by Claim 3:
There exists $x_1\in X$ with

$||x_1||_X< 2^{-1}, \ ||y-T(x_1)||_Y<2^{-1}r$

Use Claim 3 again, to pick $x_2\in X$ with

$||x_1||_X< 2^{-2}, \ ||y-T(x_1+x_2)||_Y< 2^{-2}r$

and continue inductively, to construct a sequence $x_n$ such that

$||x_n||_X < 2^{-n}, \ ||y-T(x_1+\ldots+x_n)||_Y< 2^{-n}r \ \ (1)$ .

Let
$x^*_n=\sum_{k=1}^{n} x_k.$ Then, simple calculus gives us that $(x_n^*)$ is a Cauchy sequence, and must therefore converge to an element
$x\in X$ such that

$||x||_X<\sum||x_n||_X\leq \sum2^{-n}=1$ (by the first inequality of (1))

and $||y-T(x_n^*)||_Y\leq 2^{-n}r\Rightarrow \lim T(x_n^*)=T(x)=y$ (by the second inequality of (1)). Thus $y\in T(B(0,1))$ .

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 $T^{-1}$ is also continuous.

Proof. By Claim 4, every $x\in X$ with $||T(x)||_Y< r$ satisfies $||x||_X<1$ . So $||x||_X\leq r^{-1}||Tx||_Y$ , which means that the inverse operator $T^{-1}$ is continuous.

Porisma 2. Consider a linear space $S$ , endowed with two norms $||\cdot||_1, \ ||\cdot||_2$ , such that $S$ is a banach space for either norms. If there exists $c>0$ with $||x||_2\leq c||x||_1 \ \forall x\in X$ , then there exists $c^*>0$ with $||x||_1\leq c^*||x||_2 \ \forall x\in X$ .

Proof. Apply Porisma 1 for $X=(S,||\cdot||_1)$ , $Y=(S,||\cdot||_2)$ and $T$ the identity mapping.

(Note. $c^*$ 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 $T:X\rightarrow Y$ be a linear operator between two Banach spaces. If its graph,

$G(T)=\{(x,Tx): \ x\in X\}$

is a closed subset in the space $X\times Y$ , then $T$ is continuous.

Proof.
Consider $X$ equipped with the norms

$||x||_1=||x||_X+||Tx||_Y, \ ||x||_2=||x||_X.$

Since $G(T)$ is closed, $X$ is a Banach space for either norm. We observe that $||x||_2\leq ||x||_1$ ; thus Porisma 2 implies

$||x||_1=||x||_X+||Tx||_Y\leq c||x||_2=c||x||_X$ ,

that is

$||Tx||_Y\leq (c-1)||x||_X \ \forall x\in X$ ,

so $T$ is continuous.

**miss_lolitta** · April 19th 2006, 03:23 PM

miss_lolitta is here

Rebesques thanks soooooosooooooo

much

rozana good luck

**rosana** · April 21st 2006, 11:32 PM

Rebesques thanks sooooooooooooooooooooooooo much for your caring
and helping me in this difficult theorem

hiiiiiiiiiiiiiiii miss lolita
thanks for inter this amazing theorem

**Rebesques** · April 22nd 2006, 07:53 AM

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.

**miss_lolitta** · April 22nd 2006, 04:16 PM

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

take care!!

**rosana** · April 25th 2006, 02:05 AM

hiiiiiiiiiiiiiiiiiiiiiii
I believe that every one who enter this field has a great mind like you Repesque .
miss-loltta my best friend

Good Luck

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