Let $S$ be a subset of the $\mathbb K-$normed vector space $V$ such that $$\sup_{x\in S}|T(x)|<\infty$$ for any $T\in\mathcal L(V,\mathbb K)$. Prove that $S$ is a limited subset of $V$.
Hello everyone, I found this exercise but I can't have ideas to start with, I have assumed that $S$ is not bounded, therefore for all $M>0$ exists $x_0\in S$ such that $\|x_0\|_V>M$. Or I can say that there is a sequence that converges to infinity. But since $\sup_{x\in S}|T(x)|<\infty$, then there is a sub-succession $(x_{n_j})$ such that $T(x_{n_j})$ is convergent. Then I don't know what to do. Any ideas please. I found this but I don't understand it well.
This is an application of Uniform Boundeness Principle. On the Banach space $L(V,K)$ define a family of continuous linear maps $\{F_x: x \in S\}$ by $F_x(T)=Tx$. Since $\sup \{|F_x(T)|: x \in S)\} <\infty$ for each fixed $T$ it follows by the theorem that $\sup \{|F_x(T)|: T \in L(V,K), \|T\|\leq 1, x \in S \} <\infty$. But $\sup \{|F_x(T)|: T \in L(V,K), \|T\|\leq 1 \}=\|x\|$.