# bilinear map and differentiability

let $f\colon \mathbb R^{2}\times\mathbb R^{2} \to \mathbb R$ be a bilinear map,i.e,linear in each variable separately.Then for $(V,W)$ in $\mathbb R^{2}\times \mathbb R^{2}$,the derivative $Df(V,W)$ evaluated on $(H,K)$ in $\mathbb R^{2}\times \mathbb R^{2}$ is given by
1.$f(V,K)+f(H,W)$
2.$f(H,K)$
3.$f(V,H)+f(W,K)$
4.$f(H,V)+f(W,K)$

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Please: use LaTeX to write mathematics in this site! You can go the FAQ section and follow directions. – DonAntonio Dec 19 '12 at 9:01

By bilinearity we have $$f(V+H,W+K)=f(V,W)+f(V,K)+f(H,W)+f(H,K)\ .$$ Now $f(H,K)$ can be estimated as $$|f(H,K)|\leq |H|\>|K|\ \max_{X,Y\in S^1}f(X,Y)\leq c\bigl(|H|^2+|K|^2\bigr)\ .$$ Therefore we have \eqalign{f(V+H,W+K)-f(V,W)&=f(V,K)+f(H,W)+o\left(\sqrt{|H|^2+|K|^2}\right)\cr & \qquad\bigl((H,K)\to(0,0)\bigl)\ ,\cr} and this is saying that $$Df(V,W).(H,K)=f(V,K)+f(H,W)\ .$$
Why is $\max_{X,Y \in S^1} f(X,Y)$ finite ? – Lucyfer Zedd Jan 28 at 18:55
@LucyferZedd: Because $f$ is continuous on ${\mathbb R}^2\times{\mathbb R}^2$, and $S^1\times S^1$ is compact. – Christian Blatter Jan 28 at 19:03