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 $x\in X$ ?
If $\|y\| < 1$ then $\varepsilon y = (x_0+\varepsilon y) -x_0 \in B_{\varepsilon}^X(x_0) - B_{\varepsilon}^X(x_0)$, hence $\|T_{\alpha}y\| = \dfrac{1}{\varepsilon}\|T_{\alpha}(\varepsilon y)\| \leqslant \dfrac{2n}{\varepsilon}$.