# Thread: Banach Steinhaus Theorem - Normed space

1. ## Banach Steinhaus Theorem - Normed space

Let $\displaystyle F$ be a collection of continuous linear maps from a Banach space $\displaystyle X$ into a normed space $\displaystyle Y$. Then the following are equivalent:
(1) $\displaystyle sup_{T \in F}||Tx||< \infty \forall x \in X$
(2) there exists $\displaystyle t>0$ such that $\displaystyle ||Tx|| \leq t \forall x \in X$ with $\displaystyle ||x|| \leq 1$ and $\displaystyle T \in F$
(3) there exists $\displaystyle t>0$ such that $\displaystyle ||Tx|| \leq t||x|| \forall x \in X$ and $\displaystyle T \in F$

2. Originally Posted by dori1123
Let $\displaystyle F$ be a collection of continuous linear maps from a Banach space $\displaystyle X$ into a normed space $\displaystyle Y$. Then the following are equivalent:
(1) $\displaystyle sup_{T \in F}||Tx||< \infty \forall x \in X$
(2) there exists $\displaystyle t>0$ such that $\displaystyle ||Tx|| \leq t \forall x \in X$ with $\displaystyle ||x|| \leq 1$ and $\displaystyle T \in F$
(3) there exists $\displaystyle t>0$ such that $\displaystyle ||Tx|| \leq t||x|| \forall x \in X$ and $\displaystyle T \in F$
That is the statement of the Banach–Stainhaus theorem, and you can find a proof of it here. Is that what you wanted?

3. I still don't know how to prove (3) ==> (1), can someone explain? Thank you.

4. Originally Posted by dori1123
I still don't know how to prove (3) ==> (1), can someone explain? Thank you.
$\displaystyle (3)\,\Rightarrow\,(1)$ is almost immediate. For a fixed x, $\displaystyle \sup_{T\in F}\|Tx\|\leqslant t\|x\|<\infty$.