Let $\{f_n\}$ be sequence of non zero bounded linear functionals on a Banach space X. Show that there is $x\in X$ so that $f_n(x)\ne0$, for all $n\in \Bbb N$.
I am confused, non zero functional means on set of non zero measure the functional is non zero? I have tried so idea but does not work. I am really confused which theorem from Banach space is used here. Thanks in Advance.
Assume the claim is not true. Then every $x\in X$ belongs to $\ker f_n$ for some $n$. Thus $$ X = \cup_{i=1}^n \ker f_n. $$ By Baire category theorem, one of the closed sets $\ker f_n$ has non-empty interior. Let $k$ be such that $\ker f_k$ has non-empty interior. Let $x_0\in \ker f_k$ and $\rho>0$ such that $B_\rho(x_0) \subset \ker f_k$. This ball is translated into the origin: For each $x\in B_\rho(x_0)$ the point $x-x_0\in B_\rho(0)$ is in $\ker f_k$. This proves $$ B_\rho(0)\in \ker f_k. $$ It remains to show $f_k=0$. Take $y\in X$, $y\ne 0$. Then $\frac{\rho}{2\|y\|}y\in B_\rho(0)$ and $$ f_k(y) = \frac{2\|y\|}\rho f_k\left( \frac{\rho}{2\|y\|}y\right)=0. $$ Hence, $f_k(y)=0$ for all $y\in X$, which implies $f_k=0$. A contradiction.