1
$\begingroup$

Given a vector space $V$ over $\mathbb{R}$ with a norm $||*|| $. Can $(x,y)\rightarrow(x+y)$ be an example of continous bilinear map, if yes, can you please exlain why?

Definition of continuous bilinear map $\lambda$ on $V\times V \rightarrow V$ is:

For all $v,w\in V$, there is $C>0$ such that: $||\lambda(v,w)||\le C||v||||w||$, how can I proceed from here?

$\endgroup$
  • 1
    $\begingroup$ There is something strange: $x+y \in V$. A bilinear form sends vectors to real numbers. $\endgroup$ – Siminore Apr 19 '14 at 10:07
  • $\begingroup$ Yes, sorry, it is bilinear map $\endgroup$ – L.G Apr 19 '14 at 10:14
  • 1
    $\begingroup$ It should be: there exists $C>0$ such that for all $v,w\in V$ ... not the other way around. $\endgroup$ – J.R. Apr 19 '14 at 10:20
1
$\begingroup$

Your map $\lambda:V\times V\rightarrow V$ is continuous, but not bilinear:

For $\mu\not=0\in \mathbb{R}$ and $v,w\not=0\in V$:

$$\lambda(\mu v,w)=\mu v+ w\not = \mu (v+w)=\mu\cdot\lambda(v,w)$$

However, $\lambda$ is a linear map from the vector space $V\times V$ to $V$. Therefore it is continuous if and only if there exists $C>0$ such that

$$\|\lambda(x,y)\|_V \le C \|(x,y)\|_{V\times V}\tag{1}$$

Note that $\|x\|_V \|y\|_V$ is not a norm on $V\times V$. A suitable norm (where I mean by suitable that it generates the product topology induced by $\|\cdot\|_V$) would for example be

$$\|(x,y)\|_V = \|x\|_V + \|y\|_V$$

With respect to that norm, $\lambda$ clearly satisfies $(1)$ with $C=1$ by the triangle inequality.

$\endgroup$
  • $\begingroup$ But this is for the specific norm on $V \times V$, we can have a different norm on $V$ right, how can we show continuity in general? $\endgroup$ – L.G Apr 19 '14 at 11:06
  • 1
    $\begingroup$ All norms generating the product topology are equivalent, i.e. they give the same notions of continuity. You can see continuity of $\lambda$ also directly: if $x_n\rightarrow x$, $y_n\rightarrow y$, then $x_n+y_n\rightarrow x+y$. $\endgroup$ – J.R. Apr 19 '14 at 11:26
  • $\begingroup$ Oh thats true, thank you! $\endgroup$ – L.G Apr 19 '14 at 11:32

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.