# Continuity of bilinear maps

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?

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

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.

• 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? – L.G Apr 19 '14 at 11:06
• 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 $x_n+y_n\rightarrow x+y$. – J.R. Apr 19 '14 at 11:26
• Oh thats true, thank you! – L.G Apr 19 '14 at 11:32