# 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 $kU$ 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 ProofOfOpenMappingTheorem 2013-03-22 16:23:31 2013-03-22 16:23:31 Statusx (15142) Statusx (15142) 9 Statusx (15142) Proof msc 30A99 msc 46A30