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 Answer 1


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.

  • $\begingroup$ That was what I was missing! thanks $\endgroup$
    – Johan
    Dec 5, 2012 at 7:02

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