Let $X$ be an infinite dimensional normed linear space and $x^*_1, x^*_2, \dots,x^*_k$ be k linear independent elements in the dual space $X^*$. Can we find an element in $X$ such that $x^*_1(y)<0, x^*_2(y)<0, \dots,x^*_k(y)<0$?
I know this claim fails for finite dimensional spaces and maybe we should use some similar trick as in proving $\cap^k_{i=1}ker(x^*_i)\ne 0$. Can anybody show me a proof?
Thanks in advance!
Consider the map $T : X \to \mathbb{R}^k$ defined by $Tx = (x_1^*(x), \dots, x_k^*(x))$. I claim $T$ is surjective. If not, let $(a_1, \dots, a_k)$ be a nonzero element of the orthogonal complement of the image $TX$. Then $\sum_{i=1}^k a_i x_i^*(x) = 0$ for all $x$. This means $\sum_{i=1}^k a_i x_i^* = 0$, contradicting linear independence of the $\{x_i^*\}$. So in particular, $(-1,\dots, -1) \in TX$, meaning there is an $x$ such that $Tx = (-1,\dots,-1)$, i.e. $x_i^*(x) = -1$ for every $x$.
You don't need to assume $X$ is infinite dimensional, nor use the norm. In fact, $X$ can be any vector space whatsoever. (Though when $X$ has dimension less than $k$ the claim is vacuous, because in that case you can't find $k$ linearly independent elements of $X^*$. But it's still true!)