TITLE: A detail in the proof of Banach-Steinhaus theorem that I don't understand
QUESTION [6 upvotes]: I am studying functional analysis and I have seen the Banach-Steinhaus
theorem.
For starters, the motivation given was the question about when $\{T_{\alpha}\}_{\alpha\in A}$
are bounded by $M$ (here the linear operators are $T_{\alpha}:X\to Y$).
The proof given was:

Define  $$ F_{n}:=\{x\in X\mid||T_{\alpha}x||\leq n\,\forall\alpha\} $$
Then by the assumption that the operators are pointwise bounded we get
  $$ X=\cup_{n}F_{n} $$
By Baire category theorem there is $n$ s.t $F_{n}$ contains a ball
  $B_{\epsilon}^{X}(x_{0})$.
Pick $\alpha$.  $$
 T_{\alpha}(B_{\epsilon}^{X}(x_{0})-B_{\epsilon}^{X}(x_{0}))\subseteq
 B_{2n}^{Y}(0) $$
Hence  $$ ||T_{\alpha}||\leq2n\cdot\frac{1}{\epsilon} $$

Can someone please explain the "Hence" part at the end ? 
So all $T_{\alpha}$ map some subset of $X$ to a bounded ball of
a fixed radius $2n$, how did we bound $T_{\alpha}$ for all $x\in X$
?

REPLY [5 votes]: If $\|y\| < 1$ then $\varepsilon y = (x_0+\varepsilon y) -x_0 \in B_{\varepsilon}^X(x_0) - B_{\varepsilon}^X(x_0)$, hence $\|T_{\alpha}y\| = \dfrac{1}{\varepsilon}\|T_{\alpha}(\varepsilon y)\| \leqslant \dfrac{2n}{\varepsilon}$.