[proofplan]
We expand each of the three functions in the progression average using the stated [Fourier inversion formula](/theorems/528). After collecting character factors, the sums over $x$ and $d$ become character sums over $G$. Character orthogonality forces the two dual linear constraints $r_0+r_1+r_2=0$ and $r_1+2r_2=0$, whose solutions are exactly $(r_0,r_1,r_2)=(r,-2r,r)$. Substituting this parametrisation and tracking the factors of $N$ gives the claimed factor $N^{-3}$.
[/proofplan]
[step:Record the character orthogonality identity needed for the two averages]
For $\psi\in\widehat G$, define the finite character sum
\begin{align*}
S(\psi):=\sum_{x\in G}\psi(x).
\end{align*}
Then
\begin{align*}
S(\psi)=N \quad \text{if } \psi=0.
\end{align*}
\begin{align*}
S(\psi)=0 \quad \text{if } \psi\ne 0.
\end{align*}
Indeed, if $\psi=0$, then $\psi(x)=1$ for every $x\in G$, so $S(\psi)=N$. If $\psi\ne 0$, choose $a\in G$ such that $\psi(a)\ne 1$. Translation by $a$ is a bijection from $G$ to $G$, so
\begin{align*}
S(\psi)=\sum_{x\in G}\psi(x+a).
\end{align*}
Since $\psi$ is a character, $\psi(x+a)=\psi(x)\psi(a)$ for every $x\in G$. Hence
\begin{align*}
S(\psi)=\psi(a)S(\psi).
\end{align*}
Because $\psi(a)\ne 1$, this implies $S(\psi)=0$.
[guided]
We need one finite-group Fourier fact: a character sums to zero unless it is the trivial character. Let $\psi\in\widehat G$, and define
\begin{align*}
S(\psi):=\sum_{x\in G}\psi(x).
\end{align*}
If $\psi=0$, then $\psi$ is the trivial character, so $\psi(x)=1$ for all $x\in G$. Therefore
\begin{align*}
S(\psi)=\sum_{x\in G}1=N.
\end{align*}
Now suppose $\psi\ne 0$. Since $\psi$ is not the trivial character, there exists $a\in G$ with $\psi(a)\ne 1$. The map $G\to G$ given by $x\mapsto x+a$ is a bijection, so reindexing the finite sum gives
\begin{align*}
S(\psi)=\sum_{x\in G}\psi(x+a).
\end{align*}
Because $\psi$ is a [group homomorphism](/page/Group%20Homomorphism) from the additive group $G$ to the multiplicative group $\mathbb C^\times$, we have $\psi(x+a)=\psi(x)\psi(a)$. Thus
\begin{align*}
S(\psi)=\sum_{x\in G}\psi(x)\psi(a)=\psi(a)\sum_{x\in G}\psi(x)=\psi(a)S(\psi).
\end{align*}
Since $\psi(a)\ne 1$, subtracting $\psi(a)S(\psi)$ from both sides gives
\begin{align*}
(1-\psi(a))S(\psi)=0.
\end{align*}
The scalar $1-\psi(a)$ is nonzero, hence $S(\psi)=0$. This proves the orthogonality identity.
[/guided]
[/step]
[step:Insert Fourier inversion into the three-term progression average]
By the Fourier inversion formula in the statement, for every $x,d\in G$,
\begin{align*}
f_0(x)=\frac{1}{N}\sum_{r_0\in\widehat G}\widehat f_0(r_0)r_0(x),
\end{align*}
\begin{align*}
f_1(x+d)=\frac{1}{N}\sum_{r_1\in\widehat G}\widehat f_1(r_1)r_1(x+d),
\end{align*}
and
\begin{align*}
f_2(x+2d)=\frac{1}{N}\sum_{r_2\in\widehat G}\widehat f_2(r_2)r_2(x+2d).
\end{align*}
Substituting these three finite expansions into the definition of $\Lambda_3(f_0,f_1,f_2)$ and interchanging finite sums gives
\begin{align*}
\Lambda_3(f_0,f_1,f_2)=\frac{1}{N^5}\sum_{r_0,r_1,r_2\in\widehat G}\widehat f_0(r_0)\widehat f_1(r_1)\widehat f_2(r_2)\sum_{x,d\in G}r_0(x)r_1(x+d)r_2(x+2d).
\end{align*}
[/step]
[step:Collect the character factors in the variables $x$ and $d$]
Fix $r_0,r_1,r_2\in\widehat G$. Since each $r_i$ is a character, for every $x,d\in G$,
\begin{align*}
r_1(x+d)=r_1(x)r_1(d)
\end{align*}
and
\begin{align*}
r_2(x+2d)=r_2(x)r_2(2d)=r_2(x)(2r_2)(d).
\end{align*}
Using additive notation in $\widehat G$, the product of the three $x$-factors is $(r_0+r_1+r_2)(x)$, and the product of the two $d$-factors is $(r_1+2r_2)(d)$. Hence
\begin{align*}
r_0(x)r_1(x+d)r_2(x+2d)=(r_0+r_1+r_2)(x)(r_1+2r_2)(d).
\end{align*}
Therefore
\begin{align*}
\sum_{x,d\in G}r_0(x)r_1(x+d)r_2(x+2d)=\left(\sum_{x\in G}(r_0+r_1+r_2)(x)\right)\left(\sum_{d\in G}(r_1+2r_2)(d)\right).
\end{align*}
[/step]
[step:Use orthogonality to impose the two dual linear constraints]
Applying the character orthogonality identity to the characters $r_0+r_1+r_2$ and $r_1+2r_2$, we get
\begin{align*}
\sum_{x,d\in G}r_0(x)r_1(x+d)r_2(x+2d)=N^2
\end{align*}
when
\begin{align*}
r_0+r_1+r_2=0
\end{align*}
and
\begin{align*}
r_1+2r_2=0,
\end{align*}
and the same double sum is $0$ otherwise. Consequently,
\begin{align*}
\Lambda_3(f_0,f_1,f_2)=\frac{1}{N^3}\sum_{\substack{r_0, r_1, r_2\in\widehat G, r_0+r_1+r_2=0, r_1+2r_2=0}}\widehat f_0(r_0)\widehat f_1(r_1)\widehat f_2(r_2).
\end{align*}
The normalization is $N^{-3}$ because the original average contributes $N^{-2}$, the three Fourier inversions contribute $N^{-3}$, and the two surviving character sums contribute $N^2$.
[/step]
[step:Parametrize the surviving triples and obtain the stated formula]
Let $r:=r_2\in\widehat G$. The constraint $r_1+2r_2=0$ gives
\begin{align*}
r_1=-2r.
\end{align*}
Substituting this into $r_0+r_1+r_2=0$ gives
\begin{align*}
r_0-2r+r=0,
\end{align*}
so
\begin{align*}
r_0=r.
\end{align*}
Thus the triples satisfying both constraints are exactly
\begin{align*}
(r_0,r_1,r_2)=(r,-2r,r)
\end{align*}
with $r\in\widehat G$. Substituting this parametrisation into the constrained Fourier sum yields
\begin{align*}
\Lambda_3(f_0,f_1,f_2)=\frac{1}{N^3}\sum_{r\in\widehat G}\widehat f_0(r)\widehat f_1(-2r)\widehat f_2(r).
\end{align*}
This is the desired Fourier expansion.
[/step]
