Let $X$ be an infinite dimensional Banach space. And let $X^{\mathrm{*}}$ be the space of linear continuous functionals( $f : X \rightarrow \mathbb{R}$ linear and continuous). Assume $X^{\mathrm{*}}$ is separable, and take $(x^{\mathrm{*}}_n)$for $n \in \mathbb{N}$ be a dense subset of $X^{\mathrm{*}}$. Is it true that $\bigcap _{i=0}^{\infty} \mathrm{ker}(x^{\mathrm{*}}_n)$ is trivial?
Infinite intersection of kernels
1 Answers
Consider $x\in \bigcap\limits_{n=0}^\infty\mathrm{Ker} (x_n^*)$, so $x_i^*(x)=0$ for all $n\in\mathbb{Z}_+$. Consider arbitrary functional $x^*\in X^*$. Since $\{x_n^*:n\in\mathbb{Z}_+\}$ is dense in $X^*$, then there exist some sequence $\{x_{n_k}^*:k\in\mathbb{Z}_+\}$ that is norm convergent to $x^*$, i.e. $\lim\limits_{k\to\infty}\Vert x^*-x_{n_k}^*\Vert=0$. In particular $ \Vert x^*(x)\Vert=\lim\limits_{k\to\infty}\Vert x^*(x)-x_{n_k}^*(x)\Vert\leq \left(\lim\limits_{k\to\infty}\Vert x^*-x_{n_k}^*\Vert\right)\Vert x\Vert=0 $ Thus for all $x^* \in X^*$ we have $x^*(x)=0$. By corollary of Hahn-Banach theorem $x=0$. Since $x\in \bigcap\limits_{i=0}^\infty\mathrm{Ker} (x_n^*)$ is arbitrary we conclude $ \bigcap\limits_{n=0}^\infty\mathrm{Ker} (x_n^*)=\{0\} $