Let $E, F, G$ be $3$ normed vector spaces. $B : E \times F \to G$ is a continuous bilinear form. Show that $B$ is differentiable on $E \times F$.
I need to show that: $dB_{(a_1,a_2)}(v_1,v_2) = B(a_1,v_2)+B(a_1,v_2)$
To show its continuity, can I just use the usual way: let $\big((v_1^{(n)},v_2^{(n)})\big)_n$ be a sequence in $E \times F$ which converges to $(v_1,v_2)$ and show that $\displaystyle dB_{(a_1,a_2)}(v_1^{(n)},v_2^{(n)}) \to dB_{(a_1,a_2)}(v_1,v_2)$?
(I just ask for the validity of this approach, NOT how to prove the continuity)
It seems to me, unless I'm missing something, that what you need to prove first is that for $(x,y),(h,k)\in E\times F$ you have $B(x+h,y+k)=B(x,y)+dB_{(x,y)}(h,k)+o(||h||+||k||)$ as $(h,k)\to 0$. This is the general definition of derivative, and note $dB_{(x,y)}\in L(E\times F,G)$.
Because $B$ is bilinear you'll be able to write $B(x+h,y+k)=B(x,y+k)+B(h,y+k)=B(x,y)+B(x,k)+B(h,y)+B(h,k)$. Because $B$ is continuous you should be able to bound $||B(u,v)||$ in terms of $||u||$ and $||v||$ for any $(u,v)$.
The sequence limit you've written is for sure a way of stating what you have to prove, once you've proven the existence of the derivative, in order to show that the derivative is continuous.