If $f$ is a bilinear function then prove that $Df(a,b).(x,y)=f(a,y)+f(x,b)$ Let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p$ be a bilinear function.
Prove that

$$Df(a,b).(x,y)=f(a,y)+f(x,b)$$

Here $Df(a,b)$ is the Jacobian.
My Problem
I'm not able to understand what does the dot(.) right after $Df(a,b)$ means? Is it a typo? OR does it mean something?
What I think
I think the $(x,y)$ after the dot is input to the Jacobian but
the Jacobian $(Df(a,b))$ will be A $p \times 2$ matrix and RHS is $p \times 1$. Also I'm not entirely sure that the Jacobian is $p \times 2$ matrix, if Jacobian is some combination of $f$ then that would make a lot of sense because then it will take $(x, y)$ as input and output a $p \times 1$ vector(matrix).
 A: Here $\Bbb R^n\times\Bbb R^m$ is identified with $\Bbb R^{n+m}$, and the Jacobian is a $p\times(n+m)$ matrix.
Its input $(x,y)$ is considered as a column vector $\in\Bbb R^{n+m}$ with the first $n$ coordinates coming from $x$ and the last $m$ coordinates coming from $y$.
Dot means matrix multiplication, yielding both sides of the equation to be proved an element of $\Bbb R^p$.
A: Concerning the dot: When a linear map $A$ is applied to a vector $x$ one usually writes $Ax$. But in some cases, like the one in your problem there are various nested maps at stake. Then one often writes a bottom dot between the linear map  and the vector  to which the map is applied. The map here is $Df(a,b):\>{\mathbb R}^{n+m}\to{\mathbb R}^p$, and the vector is $(x,y)\in{\mathbb R}^{n+m}$. Write $(a,b)=:c$, $\>(x,y)=:z$.
In order to find $Df(c)$ we compute
$$\eqalign{f\bigl(c+z\bigr)-f(c)&=f(a+x,b+y)-f(a,b)\cr  
&= f(a,y)+f(x,b)+f(x,y)\ ,\cr}$$
by bilinearity of $f$. The map $L:\>{\mathbb R}^{n+m}\to{\mathbb R}^p$ defined by
$$Lz:=f(a,y)+f(x,b)\qquad\bigl(z=(x,y)\in{\mathbb R}^{n+m}\bigr)$$
is linear (note that $x$ and $y$ depend linearly on $z$), and we have
$${f\bigl(c+z\bigr)-f(c)-Lz\over
|z|}={f(x,y)\over|z|}\ .\tag{1}$$
Now there is a constant $C$ such that $$|f(x,y)|\leq C|x|\,|y|\leq C|z|^2\ ,$$ and this implies $\lim_{z\to0}{f(x,y)\over|z|}=0$.
From $(1)$ we then can read off that $Df(c)=L$, according to the general definition of the derivative in a multidimensional situation.
