# proof of open mapping theorem

We prove that if $\mathrm{\Lambda}:X\to Y$ is a continuous^{} linear surjective map between Banach spaces^{}, then $\mathrm{\Lambda}$ is an open map. It suffices to show that $\mathrm{\Lambda}$ 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 $\mathrm{\Lambda}$ is surjective, $Y=\mathrm{\Lambda}(X)=\mathrm{\Lambda}({\cup}_{k\in \mathbb{N}}kU)={\cup}_{k\in \mathbb{N}}\mathrm{\Lambda}(kU)$. 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 $\mathrm{\Lambda}(kU)$.

Let ${y}_{0}\in W$, and pick $\eta >0$ so that ${y}_{0}+y\in W$ for all $y$ with $$. Then ${y}_{0}$ and ${y}_{0}+y$ are limit points^{} of $\mathrm{\Lambda}(kU)$, so there are sequences ${x}_{i}^{\prime}$ and ${x}_{i}^{\prime \prime}$ in $kU$ with $\mathrm{\Lambda}{x}_{i}^{\prime}\to {y}_{0}$ and $\mathrm{\Lambda}{x}_{i}^{\prime \prime}\to {y}_{0}+y$. Letting ${x}_{i}={x}_{i}^{\prime \prime}-{x}_{i}^{\prime}$, we have $$ and $\mathrm{\Lambda}{x}_{i}\to y$. So for any $y\in \eta V$ there is a sequence ${x}_{i}$ in $2kU$ with $\mathrm{\Lambda}{x}_{i}\to y$. Then by the linearity of $\mathrm{\Lambda}$, we have that for any $\u03f5>0$ and any $y\in Y$, there is an $x\in X$ with:

$$ and $$ (1)

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

Now let $y\in \delta V$ and $\u03f5>0$. Then there is some ${x}_{1}$ with $$ and $$. Define a sequence ${x}_{n}$ inductively as follows. Assume:

$$ (2)

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

$$ (3)

and $$, so (2) is satisfied for ${x}_{n+1}$.

Put ${s}_{n}={x}_{1}+{x}_{2}+\mathrm{\dots}+{x}_{n}$. Then from (3), ${s}_{n}$ is a Cauchy sequence^{}, and so, since $X$ is complete, it converges to some $x\in X$. By (2), $\mathrm{\Lambda}{s}_{n}\to y$, and by the continuity of $\mathrm{\Lambda}$, $\mathrm{\Lambda}{s}_{n}\to \mathrm{\Lambda}x$, so $\mathrm{\Lambda}x=y$. Also, $$. Thus $\mathrm{\Lambda}((1+\u03f5)U)\supset \eta V$, or $\mathrm{\Lambda}(U)\supset {(1+\u03f5)}^{-1}\delta V$. Since this is true for all $\u03f5>0$, we have $\mathrm{\Lambda}(U)\supset {\cup}_{\u03f5>0}{(1+\u03f5)}^{-1}\delta V=\delta V$.

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 |