How to prove that the derivatives of bent functions are balanced? Function $F:\mathbb F_2^m\rightarrow \mathbb F_2^n$ is bent iff
$$ v \cdot F$$ is bent for all nonzero $v\in \mathbb F_2^n.$
Why is this equivalent to saying $F$ is bent iff $$D_a F = F(x)+F(x+a)$$ is balanced for all nonzero $a\in \mathbb F_2^m$?
Definition:
A function $f:\mathbb F_2^m\rightarrow \mathbb F_2$ is defined to be bent
if all its Walsh-Hadamard (fourier) coefficients
$$
\hat{f}(a)=\sum_{x \in \mathbb F_2^m} (-1)^{a\cdot x + f(x)}
$$
satisfy $\vert \hat{f}(a)\vert^2=2^m.$ The property for
$f:\mathbb F_2^m\rightarrow \mathbb F_2^n$ is then defined as all nontrivial linear functions $v\cdot f(x)$ being bent.
 A: This is well understood in the context of Fourier transforms. Let $G$ and $H$ be finite abelian groups with $|H|$ divides $G$. To coincide with your definition, a function $f:G\to H$ is bent if and only if
$$\left|\sum_{x\in G}\chi(f(x))\psi(x)\right|^2=|G|,$$
for all character $\psi$ of $G$ and all nontrivial character $\chi$ of $H$. In the case where $G$ is the additive group of $\mathbb F_2^m$, note that the additive character of $\mathbb F_2^m$ is precisely $x\mapsto(-1)^{a\cdot x}$ for $a\in\mathbb F_2^m$.
We need some basic results on Fourier transforms of a function $F:G\to\mathbb C$. Let
$$\hat F(\psi)=\sum_{x\in G}F(x)\overline{\psi(x)}.$$
The Fourier inversion formula states
$$F=\frac1{|G|}\sum_{\psi}\hat F(\psi)\psi.$$
As a result,
$$\widehat{F_1*F_2}=\hat{F_1}\cdot\hat{F_2},$$
where
$$(F_1*F_2)(x)=\sum_{u\in G}F_1(x-u)F_2(u)$$
is the convolution of $F_1$ and $F_2$.
The problem boils down to the following lemmas.

Lemma 1. Let $F:G\to\mathbb C$ be a function. Then
$$|\hat F(\psi)|^2=|G|$$
for all character $\psi$ of $G$, if and only if
$$\sum_{x\in G}F(x+a)\overline{F(x)}=0$$
for all nonzero $a\in G$.


Lemma 2. Let $N:H\to\mathbb N$ be a function such that $\sum_{y\in H}N(y)=|G|$. Then $\hat N(\chi)=0$ for all nontrivial character $\chi$ of $H$, if and only if $N$ is contant.

Let $F=\chi\circ f$. Then
$$\sum_{x\in G}\chi(f(x))\overline{\psi(x)}=\hat F(\psi).$$
Let $N(y)$ be the number of $x\in G$ such that $f(x+a)-f(x)=y$. Then
$$\sum_{x\in G}F(x+a)\overline{F(x)}=\sum_{x\in G}\chi(f(x+a)-f(x))=\sum_{y\in H}N(y)\chi(y)=\hat N(\overline\chi).$$
Applying these lemmas, we obtain the desired results.
Proof of Lemma 1. (sketch) Let $F^\prime(x)=\overline{F(-x)}$. Then $\overline{\hat F(\psi)}=\hat{F^\prime}(\psi)$, and
$$|\hat F(\psi)|^2=\hat F(\psi)\hat{F^\prime}(\psi)=\widehat{F*F^\prime}(\psi).$$
By the inversion formula,
$$\sum_{x\in G}F(x+a)\overline{F(x)}=(F*F^\prime)(-a)=\frac1{|G|}\sum_{\psi}\widehat{F*F^\prime}(\psi)\cdot\psi(-a)=\frac1{|G|}\sum_{\psi}|\hat F(\psi)|^2\cdot\psi(-a).$$
The proof is done with the duality of elements and characters of $G$.
Lemma 2 is also easy by a similar argument.
