Let $\lbrace T_i \rbrace$ be a family of continuous linear operators
from Banach space $X$ to normed linear space $Y$. If for all $x \in
X$, there is an $M_x \geq 0$ for which $||T_i(x)|| \leq M_x$ for all
$T_i$, then there is a $K \geq 0$ such that for all $T_i$, $$||T_i||
\leq K.$$
I saw this proof:
For $n=1,2,3,..$, the set $A_n = \lbrace x \in X : ||T_i(x)||\leq n \rbrace$ is closed since $T_i$'s are continuous. Also, $\bigcup_{n=1}^{\infty} A_n = X$. Via Baire Category Theorem, the interior of $A_{n_0}$ is non-empty for some $n_0 \geq 1$. Picking $x_0 \in X$ and $r > 0$ such that $B(x_0,r) \subset A_{n_0}$, we have for all $i$ and for all $z \in B(0,1)$, $$||T_i(x_0 + rz)|| \leq n_0.$$ Thus, $||T_i|| \leq \frac{1}{r}(n_0 + ||T_i(x_0)||) \leq \frac{n_0}{r} + M_{x_0} = K$.
I'm somehow lost on how the centered inequality in proof was obtained. What happened to $B(x_0,r)$ that became $B(0,1)$?
Best Answer
The set of points in $B(x_0,r)$ is equivalent to the set of points of the form $x_0+zr$ where $z\in B(0,1)$. The points live in a vector space afterall.