Derivative Bilinear map I wanted to calculate the derivative of a continuous bilinear map $B: X_1 \times X_2 \rightarrow Y$. (Does anyhere know whether there is a generalisation of the notation $L(X,Y)$ that you use for the vector space of continuous linear maps to one for bilinear maps $B: X_1 \times X_2 \rightarrow Y$?)
Then we have $B(x_1 + h_1 , x_2 + h_2) = B(x_1,x_2) + B(h_1,x_2) + B(x_1,h_2) + B(h_1,h_2)$
Now I want to show that we have $\frac{B(h_1,h_2)}{|(h_1,h_2)|} \rightarrow 0$, but I do not know how.
 A: A notation I have repeatedly come across is $L^2(X_1,X_2;Y)$, with the obvious generalisation for multilinear maps on products with more factors.
If $B \colon X_1 \times X_2 \to Y$ is a continuous bilinear map between normed spaces, then you have an inequality
$$\lVert B(x,y)\rVert \leqslant \lVert B\rVert\cdot \lVert x\rVert\cdot \lVert y\rVert$$
for all $x,y$, where $\lVert B\rVert = \sup \left\lbrace \lVert B(x,y)\rVert : \lVert x\rVert \leqslant 1, \lVert y\rVert \leqslant 1\right\rbrace$ is the norm of $B$.
Then, how you show $\frac{B(h_1,h_2)}{\lvert (h_1,h_2)\rvert}\to 0$ depends on which norm you endow $X_1\times X_2$ with. If you take $\lvert (x,y)\rvert = \max \{\lVert x\rVert,\, \lVert y\rVert\}$, or $\lvert(x,y)\rvert = \lVert x\rVert + \lVert y\rVert$, or $\left(\lVert x\rVert^p + \lVert y\rVert^p\right)^{1/p}$, or something similar, then you have for $(h_1,h_2) \neq (0,0)$
$$\lVert B(h_1,h_2)\rVert \leqslant \lVert B\rVert \cdot \lVert h_1\rVert \cdot \lVert h_2\rVert \leqslant \lVert B\rVert \cdot \lvert (h_1,h_2)\rvert^2,$$
and the desired limit follows immediately. If you use a weird norm on the product, it may be a little more involved.
