4th order derivate of arbitrary function in $GF(2^8)$ At https://eprint.iacr.org/2013/455.pdf page 8, lemma 1 defines the following map from $\{0,1\}^8$ to itself:
$$
\varphi : x \mapsto \bigoplus_{a=0}^{15}{g(x \oplus a)}
$$
The paper does not explictly state the notation, so the following may be wrong, but I assume that (if a different interpretation makes more sense, feel free to correct these):


*

*$\oplus$ is field addition / vector addition (since the field is $GF(2^8)$, this is the same as component-wise xor if you represent the polynomials by the coefficents as bits)

*Perhaps then $\{0,1\}^8$ represents the vector space of dimension 8 over the field $GF(2)$, since they use $\mathbb{F}_{2^8}$ elsewhere in the paper to denote the field $GF(2^8)$?

*I have no idea what $D_1$, $D_2$, $D_4$ and $D_8$ are supposed to be
(this is their only use in the paper)


and then claims that:

The map $\varphi$ defined in Lemma 1 is a 4th-order derivative of the function $g$ (specifically $\varphi=D_1D_2D_4D_8(g)$)

How do they conclude that? There are no restrictions placed upon $g$, it is just an arbitrary map?
 A: The following is partly educated guessing, but it fits.
Basically it is just about interpreting everything carefully. Here comes the TL; DR; -version.
Assuming that:


*

*$x\in V=\Bbb{F}_2^8$ and $\phi:V\to V$,

*In the sum $x\oplus a$ the integer $a$ is transformed to an element of $V$ by using the base-2 digits as components. So if $a=\sum_{i=0}^7a_i2^i, a_i\in\{0,1\}$, and $x=(x_0,x_1,\ldots,x_7)\in V$, then
$$x\oplus a=(x_0+a_0,x_1+a_1,\ldots,x_7+a_7),$$ where here the sums of the components are sums in $\Bbb{F}_2$.

*If $n\in\{0,1,2,\ldots,7\}$ the operator $D_{2^n}$ is defined as follows. If $g:V\to V$, then $D_{2^n}(g):V\to V$ is the function defined by
$$D_{2^n}(g):x\mapsto g(x\oplus 2^n)-g(x).$$
Of course, in $V$ subtraction = addition, so instead of a difference we can use a sum just as well.


With this definition we do get the formula 
$$
D_1D_2D_4D_8(g):x\mapsto \sum_{a=0}^{15}g(x\oplus a).
$$
This is just a manifestation of the fact that, when viewed as above, the set
$W=\{0,1,2,\ldots,15\}$ is a subgroup of $V$. It is actually a subspace spanned by the linearly independent elements $1$, $2$, $4$ and $8$.
To see this we first observe that
$$
D_1(g):x\mapsto g(x\oplus1)+g(x)=\sum_{a=0}^1g(x\oplus a).
$$
Then
$$
\begin{aligned}
D_1D_2(g):x&\mapsto D_2(g)(x\oplus1)+D_2(g)(x)\\
&=\left(g((x\oplus1)\oplus2)+g(x\oplus1)\right)+\left(g(x\oplus2)+g(x)\right)\\
&=g(x\oplus3)+g(x\oplus1)+g(x\oplus2)+g(x)\\
&=\sum_{a=0}^3g(x\oplus a).
\end{aligned}
$$
Repeating the dose
$$
\begin{aligned}
D_1D_2D_4(g):x&\mapsto \sum_{a=0}^3D_4(g)(x\oplus a)\\
&=\sum_{a=0}^3\big(g(x\oplus a\oplus4)+g(x\oplus a)\big)\\
&=\sum_{a=0}^7g(x\oplus a).
\end{aligned}
$$
And one more repetition then gives the claim.
To understand why $D_{2^n}$ are called derivations we need to break the functions into components. Any function $g:V\to V$ can be written using the
components
$$
g(x)=(g_0(x),g_1(x),\ldots,g_7(x)),
$$
where the components $g_i:V\to\Bbb{F}_2$ are all boolean functions in the variables $x_0,x_1,\ldots,x_7$. Any such boolean function can be uniquely written as a polynomial in the unknowns $x_i$ ranging over $\Bbb{F}_2$. Remember that we only need to use terms $x_{i_1}x_{i_2}\cdots x_{i_k}$ with
$i_1<i_2<\cdots<i_k$ because the polynomials $x_i^2$ and $x_i$ give the same boolean function.
We obviously have
$$
g_i(x\oplus 2^n)=g_i(x_0,x_1,\ldots,x_{n-1},x_n+1,x_{n+1},\ldots,x_7).\tag{1}
$$ 
The following observation is the key:
$$
D_{2^n}(g_i)(x)=\frac{\partial g_i}{\partial x_n}(x).\tag{2}
$$
Everything in sight is linear, so it suffices to prove $(2)$ for each monomial $g_i(x)=x_{i_1}x_{i_2}\cdots x_{i_k}$. Here if $n$ does not appear among the indices $i_1,\ldots, i_k$ we have, by $(1)$, $g_i(x\oplus2^n)=g_i(x)$ for all $x\in V$. In this case we have the constant function zero on both sides of $(2)$. 
On the other hand, if $i_t=n$, then
$$
\begin{aligned}
D_{2^n}(g_i)(x)&=g_i(x\oplus2^n)+g_i(x)\\
&=g_i(x_0,x_1,\ldots,x_n+1,\ldots,x_7)+g_i(x_0,\ldots,x_7)\\
&=x_{i_1}\cdots (x_{i_t}+1)\cdots x_{i_k}+x_{i_1}x_{i_2}\cdots x_{i_k}\\
&=x_{i_1}\cdots x_{i_{t-1}} x_{i_{t+1}}\cdots x_{i_k},
\end{aligned}
$$
so $(2)$ holds in this case as well. Q.E.D.
This then implies the last claim in the form
$$
D_1D_2D_4D_8(g)=\left(\frac{\partial^4g_i}{\partial x_0\partial x_1\partial x_2\partial x_3}\right)_{i=0}^7.
$$
