proof of Banach-Steinhaus theorem

Let

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

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

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

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 $T\in ℱ$, 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 $T\in ℱ$.

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