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This question is more conceptual than other thing.

We know that if $f:U\subset \mathbb{R}^n \to \mathbb{R}^m$, where $U$ is a open subset of $\mathbb{R}^n$, is differentiable, then the derivative of $f$ in $a \in U$ is a linear transformation. That is $f' \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$, the set of the linear transformations between $\mathbb{R}^n$ and $\mathbb{R}^m$.

So, my doubt is about one bilinear map. All know that if $B: \mathbb{R}^n \hbox{ x } \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear map, we have that

$$B((u,v)+(s,t)) = B(u + s,v + t) = B(u,v) + B(u,t) + B(s,v) + B(s,t)$$

where, with some calculation, we can show that $\lim_{|(s,t)| \to 0} \frac{B(s,t)}{|(s,t)|} = 0$. In this case, we can claim that

$$B'(u,v)(s,t) = B(u,t) + B(s,v) $$

In this case, can I claim that $B' \in \mathcal{L}(\mathbb{R}^n \hbox{ x }\mathbb{R}^m ,\mathbb{R}^p) $?

This is very strange. So, what is the nature of $B'$?

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  • $\begingroup$ $B$ is regarded as a map $\mathbb R^{n+m}\to\mathbb R^p$. So its derivative is the same as a usual function. $\endgroup$ – Eclipse Sun May 31 '16 at 5:43
  • $\begingroup$ @EclipseSun I think the question is what form does the derivative have. $\endgroup$ – Tim kinsella May 31 '16 at 5:44
  • $\begingroup$ I think that the form is well know, but his structure is unknown. $\endgroup$ – orrillo May 31 '16 at 5:52
  • $\begingroup$ It seems to me the derivative has the form $$ D_{v,w}= \left( \begin{array}{ccc} M(w) & N(v) \end{array} \right)$$ where $M$ is an $p \times m$ matrix depending linearly on $w$, and $N$ is a $p\times n$ matrix depending linearly on $w$. It might also be fruitful to think about the induced map on the tensor product $\mathbb{R}^m\otimes \mathbb{R}^n$/. $\endgroup$ – Tim kinsella May 31 '16 at 5:52
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The differential $df(p,\delta p)$ of a function $f:U\to \mathbb{R}^m$ is a function of both a point $p\in U$ and of a tangent vector $\delta p\in T_pU$ at $p$.

As you note, the differential is always linear in the tangent vector; this is a defining property of the differential. Generically $df$ is an arbitrary nonlinear function of $p$, though. Some special cases are where $f$ is linear, in which case $df$ is constant in $p$, and the case where $f$ is quadratic, in which case $df$ is bilinear in $p$ and $\delta p$ (again, linearity in $\delta p$ alone is automatic).

The case of your bilinear map $B$ is just a special case of a quadratic function on $\mathbb{R}^n\times \mathbb{R}^m$. So $dB(p,\delta p)$ is bilinear in $p$ and $\delta p$, which is clear from the expression you derived for $dB$.

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