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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?

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  • $\begingroup$ 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, 2014 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, 2014 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, 2014 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, 2014 at 18:20
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    $\begingroup$ @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, 2019 at 11:08

1 Answer 1

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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.

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  • $\begingroup$ So this also assumes we have the norm $\| (v,w) \| := \max \{ \|v\| , \|w\| \}$ on $V \times V$, right? $\endgroup$
    – Adam
    Jan 29, 2020 at 20:22
  • $\begingroup$ Yes, although with slight adjustments it will work for any equivalent norm on the product space. $\endgroup$
    – Marcin Łoś
    Jan 29, 2020 at 22:19

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