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[parent] proof of Banach-Steinhaus theorem (Proof)

Let

$\displaystyle E_n = \{x\in X: \Vert T(x)\Vert\leq n\textnormal{ for all }T\in \mathcal{F}\}.$
From the hypothesis, we have that
$\displaystyle \bigcup_{n=1}^\infty E_n = X.$
Also, each $ E_n$ is closed, since it can be written as
$\displaystyle E_n = \bigcap_{T\in\mathcal{F}}{T^{-1}(B(0,n))},$
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 $ \Vert x\Vert\leq r$, we have
$\displaystyle \Vert T(x)\Vert-\Vert T(x_0)\Vert\leq \Vert T(x_0)+T(x)\Vert=\Vert T(x_0+x)\Vert\leq n$
for each $ T\in \mathcal{F}$, and so
$\displaystyle \Vert T(x)\Vert\leq n+\Vert T(x_0)\Vert$
so if $ \Vert x\Vert\leq 1$, we have
$\displaystyle \Vert T(x)\Vert= \frac{1}{r}\Vert T(rx)\Vert \leq \frac{1}{r}\left(n+\Vert T(x_0)\Vert\right) = c,$
and this means that
$\displaystyle \Vert T\Vert = \sup\{\Vert Tx\Vert: \Vert x\Vert\leq 1\} \leq c$
for all $ T\in\mathcal{F}$.



"proof of Banach-Steinhaus theorem" is owned by Koro.
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Cross-references: interior, implies, Baire's category theorem, Banach space, operators, intersection, radius, closed ball, closed, hypothesis

This is version 3 of proof of Banach-Steinhaus theorem, born on 2004-11-12, modified 2006-06-28.
Object id is 6470, canonical name is ProofOfBanachSteinhausTheorem.
Accessed 2334 times total.

Classification:
AMS MSC: 46B99 (Functional analysis :: Normed linear spaces and Banach spaces; Banach lattices :: Miscellaneous)
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