5
$\begingroup$

I've just proved the fact that every linear function on a finite dimensional normed vector space is uniformly continuous. Because, let $T:U\to V$ be a linear function on $U$ with basis : $(u_1,u_2,\dots,u_n)$ and suppose that $\|\cdot\|_u$, $\|\cdot\|_v$ be the respective norms associated with $U$ and $V$. Setting $M= \max(\|Tu_1\|_v,\dots,\|Tu_n\|_v)$, we have $$\|Tu\|_v=\|a_1Tu_1 + a_2Tu_2 +\dots+ a_nTu_n\|_v \le |a_1|\cdot\|Tu_1\|_v+\dots+|a_n|\cdot\|Tu_n\|_v \le M\|u\|_1,$$ where $\|u\|_1=|a_1|+\dots+|a_n|$. Since all norms on $U$ are equivalent, we have $\|Tu\|_v\le C\|u\|_u$ for some $C>0$. Thus, $\|Tx-Ty\| \le C\|x-y\|$ for all $x$, $y$ in $U$, thereby satisfying Lipschitz' condition.
My questions are:

  1. Is it true that a linear function from one vector space to another is always continuous? (finite dimension not assumed)
  2. If so, or else, is a continuous linear map always uniformly continuous?
$\endgroup$
2
  • 1
    $\begingroup$ For the first question, you could have a look here: Discontinuous linear functional $\endgroup$ Commented May 19, 2012 at 6:00
  • $\begingroup$ BTW Your post was edited using TeX markup - for better readability. I think learning basics of TeX is not that difficult, see e.g. this meta thread. BTW I'd think that $\|\cdot\|_U$ would be slightly better notation than $\|\cdot\|_u$, but I tried to stick with your original version as much as possible, when editing. $\endgroup$ Commented May 19, 2012 at 6:04

2 Answers 2

5
$\begingroup$

You don't need the Axiom of Choice.

Take the space of polynomials in one variable, with the norm $||p|| = \sup_{t\in [0,1]}|p(t)|$. Then take the sequence of polynomials $p_n$ defined by $p_n(t) = (\frac{x}{2})^n$. Clearly $p_n \rightarrow 0$ in this norm, but the linear functional $\phi(p) = p(4)$ is unbounded (in fact, $||p_n|| = \frac{1}{2^n}, \phi(p_n) = 2^n)$.

$\endgroup$
1
  • $\begingroup$ You are right there. I was too quick. $\endgroup$
    – abatkai
    Commented May 19, 2012 at 10:01
3
$\begingroup$

(1): no. There is no direct construction, but assuming the Axiom of Choice, it is possible to construct such a map. You can easily construct non-continuous maps which are not everywhere defined, but only densely: differential operators are typical canonical examples.

(2): yes. A continuous map is continuous at zero, hence bounded, hence Lipschitz as can be seen from your proof.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .