8
$\begingroup$

In my numeric script there is a unproved theorem, saying that a bilinear form $a \colon V\times V \to \mathbb{R}$ on a normed vector space $V$ is continuous if and only if

$$|a(v,w)| \leq c \, \|v\| \, \|w\|$$ holds for all $v,w\in V$ for some $c > 0$.

My first question is: what is ment by a continuous bilinearform? Is it according to the norm $\| (v,w) \| := \max \{ \|v\| , \|w\| \}$ (which is equivalent to $ \|(v,w)\| = \|v\| + \|w\|$) ?

If so, then I agree that such a bilinear form is continuous but I don't see that a continuous bilinear form is bounded as above. Can anyone explain this to me?

$\endgroup$
  • $\begingroup$ If $a$ is continuous, there is an $\delta$-neighborhood of the origin on which $a$ is bounded by $1$. Rescaling, we can see $a$ is bounded on unit ball around the origin. Can you see how that implies boundedness as in your question? $\endgroup$ – Marcin Łoś May 25 '14 at 17:53
  • $\begingroup$ @MarcinŁoś If I could show that a is bounded by 1 in an $\delta$-neigborhood, than I belive it is easy to show that a is bounded by 1 on the whole space $V$. Because rescaling $v$ and $w$ on both sides just cancels out. Therefore, I dont think it is true that a is bounded by 1 near the $\delta$-neigborhood. $\endgroup$ – Adam May 25 '14 at 18:16
  • $\begingroup$ How does rescaling "on both sides" cancels out? $a(\alpha u,\beta v) = \alpha\beta\, a(u, v)$. $\endgroup$ – Marcin Łoś May 25 '14 at 18:18
  • $\begingroup$ @MarcinŁoś $$|a (\alpha v ,\beta w)| \leq c || \alpha v|| ||\beta w||$$ is equivalent to $$|a (v ,w)| \leq c || v|| || w|| $$. $\endgroup$ – Adam May 25 '14 at 18:20
  • 1
    $\begingroup$ @TheCodingWombat in your exersice sheet your vector space is of finite dimension. Every linear map on a finite space is continuous. $\endgroup$ – Adam May 7 at 11:08
14
$\begingroup$

Assume $a$ is continuous at the origint. Since $a(0,0)=0$, by definition of continuity, there exists some $\delta>0$ such that $\left|a(u,v)\right|<1$ for any $\|u\|,\|v\|\leq\delta$ (here I assume maximum norm). Thus, for any $u,v$ we have $$ \left|a(u,v)\right|= \left|a\left(\frac{\|u\|}{\delta}\,\frac{\delta u}{\|u\|}, \frac{\|v\|}{\delta}\,\frac{\delta v}{\|v\|}\right)\right|= \delta^{-2}\|u\|\|v\| \left|a\left(\frac{\delta u}{\|u\|}, \frac{\delta v}{\|v\|}\right)\right|<\delta^{-2}\|u\|\|v\| $$ since $\delta\,u/\|u\|$, $\delta\, v/\|v\| \leq \delta$. Hence, $a$ is bounded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.