Let $$E_n = \{x\in X: \|T(x)\|\leq n\textnormal{ for all }T\in \mathcal{F}\}.$$ From the hypothesis, we have that $$\bigcup_{n=1}^\infty E_n = X.$$ Also, each $E_n$ is closed, since it can be written as $$E_n = \bigcap_{T\in\mathcal{F}}{T^{-1}(B(0,n))},$$ where $B(0,n)$ is the closed ball centered at $0$ with radius $n$ in $Y$ , and each of the sets in the intersection is closed due to the continuity of the operators. Now since $X$ is a Banach space, Baire's category theorem implies that there exists $n$ such that $E_n$ has nonempty interior. So there is $x_0\in E_n$ and $r>0$ such that $B(x_0,r)\subset E_n$ . Thus if $\|x\|\leq r$ , we have $$\|T(x)\|-\|T(x_0)\|\leq \|T(x_0)+T(x)\|=\|T(x_0+x)\|\leq n$$ for each $T\in \mathcal{F}$ , and so $$\|T(x)\|\leq n+\|T(x_0)\|$$ so if $\|x\|\leq 1$ , we have $$\|T(x)\|= \frac{1}{r}\|T(rx)\| \leq \frac{1}{r}\left(n+\|T(x_0)\|\right) = c,$$ and this means that $$\|T\| = \sup\{\|Tx\|: \|x\|\leq 1\} \leq c$$ for all $T\in\mathcal{F}$ .