# proof of Banach-Steinhaus theorem

Let

$${E}_{n}=\{x\in X:\parallel T(x)\parallel \le n\text{for all}T\in \mathcal{F}\}.$$ |

From the hypothesis^{}, we have that

$$\bigcup _{n=1}^{\mathrm{\infty}}{E}_{n}=X.$$ |

Also, each ${E}_{n}$ is closed, since it can be written as

$${E}_{n}=\bigcap _{T\in \mathcal{F}}{T}^{-1}(B(0,n)),$$ |

where $B(0,n)$ is the closed ball centered at $0$ with radius $n$ in $Y$,
and each of the sets in the intersection^{} is closed due to the continuity of the operators.
Now since $X$ is a Banach space^{}, Baire’s category theorem^{}
implies that there exists $n$ such that ${E}_{n}$ has
nonempty interior. So there is ${x}_{0}\in {E}_{n}$ and $r>0$ such
that $B({x}_{0},r)\subset {E}_{n}$. Thus if $\parallel x\parallel \le r$, we have

$$\parallel T(x)\parallel -\parallel T({x}_{0})\parallel \le \parallel T({x}_{0})+T(x)\parallel =\parallel T({x}_{0}+x)\parallel \le n$$ |

for each $T\in \mathcal{F}$, and so

$$\parallel T(x)\parallel \le n+\parallel T({x}_{0})\parallel $$ |

so if $\parallel x\parallel \le 1$, we have

$$\parallel T(x)\parallel =\frac{1}{r}\parallel T(rx)\parallel \le \frac{1}{r}\left(n+\parallel T({x}_{0})\parallel \right)=c,$$ |

and this means that

$$\parallel T\parallel =sup\{\parallel Tx\parallel :\parallel x\parallel \le 1\}\le c$$ |

for all $T\in \mathcal{F}$.

Title | proof of Banach-Steinhaus theorem |
---|---|

Canonical name | ProofOfBanachSteinhausTheorem |

Date of creation | 2013-03-22 14:48:41 |

Last modified on | 2013-03-22 14:48:41 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 6 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 46B99 |