The derivative at any point $(u,v)\in V\times W$ is the linear map defined by $(x,y)\longmapsto B(u,y)+B(x,v)$ . I am trying to prove this using the limit definition. My limit ended up looking like ${{\displaystyle {\lim_{(h_{1},h_{2})\to0}}}}\frac{|B(h_{1},h_{2})|}{|(h_{1},h_{2})|}$ and I am trying to prove that it is $0$ , but I don't now how.
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For $(u,v) \in U \times V$ you have $$\begin{aligned} \Vert B(x+u,y+v)-B(x,y) -(B(u,y)+B(x,v)) \Vert &= \Vert B(u,v) \Vert\\ &\le \Vert B \Vert \Vert u \Vert \Vert v \Vert\\ &\le \frac{\Vert B \Vert}{2} (\Vert u \Vert + \Vert v \Vert)^2 \end{aligned}$$ Hence $$\lim\limits_{(u,v) \to (0,0)} \frac{\Vert B(x+u,y+v)-B(x,y) -(B(u,y)+B(x,v)) \Vert}{\Vert (u,v) \Vert} = 0$$
as $\Vert u \Vert + \Vert v \Vert$ is a norm on $U \times V$, and on finite dimensional spaces all norms are equivalent.
This proves that $(u,v) \mapsto B(u,y)+B(x,v)$ is the derivative of $B$ at $(x,y)$.
mathcounterexamples.net
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1How can one show that first inequality? – Eigenfield Nov 29 '16 at 11:09
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$x^2+y^2-2xy = (x-y)^2 \ge 0 $ – mathcounterexamples.net Nov 29 '16 at 11:58
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1No I'm talking about the boundlessness property. Does every bilinear map on finite dimensional spaces have that property? – Eigenfield Nov 29 '16 at 12:16
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For sure $n$-linear maps are continuous on finite dimensional spaces. Proof is similar to the one for linear map. – mathcounterexamples.net Nov 29 '16 at 12:31
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It isn't clear to me as well why every bilinear map on a finite dimensional space is bounded. – solrak Jan 29 '20 at 19:56
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@solrak You just need to apply twice what is done for a linear map. Here for more details. – mathcounterexamples.net Jan 29 '20 at 20:07