2
$\begingroup$

Let $X$ be an infinte-dimensional normed space. Let $\ell_1,\ldots, \ell_n$ be continuous linear functionals on $X$ and consider the set $U = \{x\in X : |\ell_j(x)| < 1,\;\; 1\leq j \leq n\}.$ Show that the set $U$ is unbounded.

If we put the functionals together to a linear map $L:X\rightarrow \mathbb{C}$, $L(x) = (\ell_1(x), \ell_2(x),\ldots,\ell_n(x))$. The kernel of this map is a subspace, right? How would this kernel look? would it be anything else then $x= \textbf{0}$? I seem to be stuck here.

1 Answers 1

3

Recall that if we have a linear map $T:V\rightarrow W$ where $V,W$ are vector spaces, then:

$V/\ker(T)\cong \textrm{im}(T). $ As the image of the map $L$ in question is a subspace of the finite-dimensional space $\mathbb{C}^n$, we can conclude that the kernel of $L$ is nontrivial (and is in fact infinite dimensional).

This is enough to complete the proof that $U$ is unbounded.

  • 0
    That was what I was missing! thanks2012-12-05