# proof of Banach-Steinhaus theorem

Let

 $E_{n}=\{x\in X:\|T(x)\|\leq n\textnormal{ for all }T\in\mathcal{F}\}.$

From the hypothesis, we have that

 $\bigcup_{n=1}^{\infty}E_{n}=X.$

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 $\|x\|\leq r$, we have

 $\|T(x)\|-\|T(x_{0})\|\leq\|T(x_{0})+T(x)\|=\|T(x_{0}+x)\|\leq n$

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

 $\|T(x)\|\leq n+\|T(x_{0})\|$

so if $\|x\|\leq 1$, we have

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

and this means that

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

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

Title proof of Banach-Steinhaus theorem ProofOfBanachSteinhausTheorem 2013-03-22 14:48:41 2013-03-22 14:48:41 Koro (127) Koro (127) 6 Koro (127) Proof msc 46B99