proof of open mapping theorem

We prove that if $\Lambda:X\rightarrow Y$ is a continuous linear surjective map between Banach spaces, then $\Lambda$ is an open map. It suffices to show that $\Lambda$ maps the open unit ball in $X$ to a neighborhood of the origin of $Y$ .

Let $U$ , $V$ be the open unit balls in $X$ , $Y$ respectively. Then $X=\cup_{k\in\mathbb{N}}kU$ , so, since $\Lambda$ is surjective, $Y=\Lambda(X)=\Lambda(\cup_{k\in\mathbb{N}}kU)=\cup_{k\in\mathbb{N}}\Lambda(kU)$ . By the Baire category theorem, $Y$ is not the union of countably many nowhere dense sets, so there is some $k\in\mathbb{N}$ and some open set $W\subset Y$ such that $W$ is contained in the closure of $\Lambda(kU)$ .

Let $y_{0}\in W$ , and pick $\eta>0$ so that $y_{0}+y\in W$ for all $y$ with $||y||<\eta$ . Then $y_{0}$ and $y_{0}+y$ are limit points of $\Lambda(kU)$ , so there are sequences ${x_{i}^{\prime}}$ and ${x_{i}^{\prime\prime}}$ in $k U$ with $\Lambda x_{i}^{\prime}\rightarrow y_{0}$ and $\Lambda x_{i}^{\prime\prime}\rightarrow y_{0}+y$ . Letting $x_{i}=x_{i}^{\prime\prime}-x_{i}^{\prime}$ , we have $||x_{i}||<2k$ and $\Lambda x_{i}\rightarrow y$ . So for any $y\in\eta V$ there is a sequence ${x_{i}}$ in $2kU$ with $\Lambda x_{i}\rightarrow y$ . Then by the linearity of $\Lambda$ , we have that for any $\epsilon>0$ and any $y\in Y$ , there is an $x\in X$ with:

$||x||<\delta^{-1}||y||$ and $||\Lambda x-y||<\epsilon$ (1)

where $\delta=\eta/2k$ .

Now let $y\in\delta V$ and $\epsilon>0$ . Then there is some $x_{1}$ with $||x_{1}||<1$ and $||y-\Lambda x_{1}||<\epsilon\delta$ . Define a sequence ${x_{n}}$ inductively as follows. Assume:

$||y-\Lambda(x_{1}+x_{2}+...+x_{n})||<\epsilon\delta 2^{-n}$ (2)

Then by (1) we can pick $x_{n+1}$ so that:

$||x_{n+1}||<\epsilon 2^{-n}$ (3)

and $||y-\Lambda(x_{1}+x_{2}+...+x_{n})-\Lambda(x_{n+1})||<\epsilon\delta 2^{-(n+1)}$ , so (2) is satisfied for $x_{n+1}$ .

Put $s_{n}=x_{1}+x_{2}+...+x_{n}$ . Then from (3), $s_{n}$ is a Cauchy sequence, and so, since $X$ is complete, it converges to some $x\in X$ . By (2), $\Lambda s_{n}\rightarrow y$ , and by the continuity of $\Lambda$ , $\Lambda s_{n}\rightarrow\Lambda x$ , so $\Lambda x=y$ . Also, $||x||=\lim_{n\rightarrow\infty}||s_{n}||\leq\sum_{n=1}^{\infty}||x_{n}||<1+\epsilon$ . Thus $\Lambda((1+\epsilon)U)\supset\eta V$ , or $\Lambda(U)\supset(1+\epsilon)^{-1}\delta V$ . Since this is true for all $\epsilon>0$ , we have $\Lambda(U)\supset\cup_{\epsilon>0}(1+\epsilon)^{-1}\delta V=\delta V$ .

Title	proof of open mapping theorem
Canonical name	ProofOfOpenMappingTheorem
Date of creation	2013-03-22 16:23:31
Last modified on	2013-03-22 16:23:31
Owner	Statusx (15142)
Last modified by	Statusx (15142)
Numerical id	9
Author	Statusx (15142)
Entry type	Proof
Classification	msc 30A99
Classification	msc 46A30