# quotients of Banach spaces by closed subspaces are Banach spaces under the quotient norm

Theorem - Let $X$ be a Banach space^{} and $M$ a closed subspace. Then $X/M$ with the quotient norm is a
Banach space.

Proof : In to prove that $X/M$ is a Banach space it is enough to prove that every series in $X/M$ that converges absolutely also converges in $X/M$.

Let ${\sum}_{n}{X}_{n}$ be an absolutely convergent series in $X/M$, i.e., $$. By definition of the quotient norm, there exists ${x}_{n}\in {X}_{n}$ such that

$$\parallel {x}_{n}\parallel \le {\parallel {X}_{n}\parallel}_{X/M}+{2}^{-n}$$ |

It is clear that $$ and so, as $X$ is a Banach space, ${\sum}_{n}{x}_{n}$ is convergent.

Let $x={\sum}_{n}{x}_{n}$ and ${s}_{k}={\sum}_{n=1}^{k}{x}_{n}$. We have that

$$x-{s}_{k}+M=(x+M)-({s}_{k}+M)=(x+M)-\sum _{n=1}^{k}({x}_{n}+M)=(x+M)-\sum _{n=1}^{k}{X}_{n}$$ |

Since ${\parallel x-{s}_{k}+M\parallel}_{X/M}\le \parallel x-{s}_{k}\parallel \u27f60$ we see that ${\sum}_{n}{X}_{n}$ converges in $X/M$ to $x+M$. $\mathrm{\square}$

Title | quotients of Banach spaces by closed subspaces are Banach spaces under the quotient norm |
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Canonical name | QuotientsOfBanachSpacesByClosedSubspacesAreBanachSpacesUnderTheQuotientNorm |

Date of creation | 2013-03-22 17:23:01 |

Last modified on | 2013-03-22 17:23:01 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 7 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46B99 |