Linear functional in Banach space

Question

Let $X$ be a Banach space, $(f_n)\in X^{*}$ a sequence with $f_n\neq 0 $ $ \forall n\in \Bbb N$. Show that there is a $x\in X$ such that $f_n(x)\neq 0 $ $\forall n$.

Need some help. Thank you!

Answer 1

It's something with the kernel of the functionals,right?

Right. What you want is an $x$ with $f_n(x) \neq 0$ for all $n$, so

$$x \in \bigcap_{n\in\mathbb{N}} \left(X\setminus \ker f_n\right) = X \setminus \bigcup_{n\in\mathbb{N}} \ker f_n.$$

The $f_n$ are continuous, so $\ker f_n$ is closed. $f_n \neq 0$, hence $\ker f_n \neq X$, and therefore $(\ker f_n)^{\Large\circ} = \varnothing$. $X$ is a Banach space, hence a complete metric space, hence Baire's theorem ...