Uniform Boundedness Principle

Theorem

Let $X$ be a Banach space, Y be a normed vector space. Let $\{T_{\alpha }:X\to Y{\}}_{\alpha \in \Lambda }$ be a family of bounded linear operators from $X$ to $Y$. Suppose $\forall x\in X$, ${\sup }_{\alpha \in \Lambda }\Vert T_{\alpha }x{\Vert }_Y<\infty$ i.e. for all$x\in X$, $\exists C_x<\infty$ such that $\Vert T_{\alpha }x{\Vert }_Y\le C_x$ for all $\alpha \in \Lambda$. Then ${\sup }_{\alpha \in \Lambda }\Vert T_{\alpha }{\Vert }_{X\to Y}<\infty$ i.e. $\exists C<\infty$ such that $\Vert T_{\alpha }x{\Vert }_Y\le C\Vert x{\Vert }_x$ for all $\alpha \in \Lambda$, $x\in X$.

Proof $\forall n\ge 1$, let $F_n=\{x\in X:\Vert T_{\alpha }x{\Vert }_Y\le n\forall \alpha \in \Lambda \}$, $X={\cup }_{n=1}^{\infty }{F}_n$ and $F_n={\bigcap }_{\alpha \in \Lambda }{T}_{\alpha }^{-1}\{y\in Y:\Vert y{\Vert }_{\Psi }\le n\},$ so $F_n$ is closed from continuity of $T_{\alpha }\forall \alpha \in \Lambda$ .From corollary of Baire Category Theorem, there exits some $F_m$ that contains an interior point, i.e. $B(x_0,ϵ)\subset F_m$ for some $x_0\in X$, $ϵ>0$, with $F_m$ closed, $\overline{B_{\epsilon }(x_0)}:=\left\{x\in X:\Vert x-x_0\Vert \le \epsilon \right\}\subseteq F_m.$ Let $u\in X$ with $\Vert u\Vert \le 1$ and $\alpha \in \Lambda$. Then: $\Vert T_{\alpha }(u){\Vert }_Y={\epsilon }^{-1}\Vert T_{\alpha }(x_0+\epsilon u)-T_{\alpha }(x_0){\Vert }_Y\le {\epsilon }^{-1}\left(\Vert T_{\alpha }(x_0+\epsilon u){\Vert }_Y+\Vert T_{\alpha }(x_0){\Vert }_Y\right)\le {\epsilon }^{-1}(m+m).$ Taking the supremum over $u$ in the unit ball of $X$ and over $T_{\alpha }$ it follows that ${\sup }_{\alpha \in \Lambda }\Vert T_{\alpha }{\Vert }_{\mathcal{B}(X,Y)}\le 2{\epsilon }^{-1}m<\infty .$