When I read the artical A new proof of Roth's theorem on arithmetic progressions, I met an analystic inequality: $$|\Lambda(f)-\Lambda(g)|\leq|\widehat{f}(r)-\widehat{g}(r)|.$$
Here $f,g:\mathbb{F}_p\to\left\{0,1\right\}$, $h:\mathbb{F}_p\to\left\{0,\frac{p}{3}\right\}$ with$$g(n)=(f*h)(n)=\frac{f(n)+f(n-x)+f(n-2x)}{3},$$ here $x$ is a postive real number.
And $$\Lambda(f)=p^{-2}\sum_{x,d\in\mathbb{F}_p}f(x)f(x+d)f(x+2d)=\sum_{r\in\mathbb{F}_p}\widehat{f}(r)^2\widehat{f}(-2r),$$ where $\widehat{f}(r)$ is the $r$th fourier coefficient of $f$.
My attempt:
$\begin{align*} LHS=|\Lambda(f)-\Lambda(g)|&=\left|\sum_{r\in\mathbb{F}_p}\widehat{f}(r)^2\widehat{f}(-2r)-\sum_{r\in\mathbb{F}_p}\widehat{f}(r)^2\widehat{f}(-2r)\widehat{h}(r)^2\widehat{h}(-2r)\right|\\ &=\left|\sum_{r\in\mathbb{F_p}}\left(\widehat{f}(r)^2\widehat{f}(-2r)\right)\left(1-\widehat{h}(r)^2\widehat{h}(-2r)\right)\right|\\ &\leq\sqrt{\left(\sum_{r\in\mathbb{F}_p}\widehat{f}(r)^4\widehat{f}(-2r)^2\right)\cdot\left(\sum_{r\in\mathbb{F}_p}\left(1-\widehat{h}(r)^2\widehat{h}(-2r)\right)^2\right)} \end{align*}$
And
$\begin{align*} RHS=|\widehat{f}(r)-\widehat{g}(r)|&=\left|\widehat{f}(r)-\widehat{f}(r)\cdot\widehat{h}(r)\right|\\ &=\left|\widehat{f}(r)\left(1-\widehat{h}(r)\right)\right|\\ &\geq\frac{1}{p}\left|\sum_{r\in\mathbb{F}_p}\widehat{f}(r)\left(1-\widehat{h}(r)\right)\right| \end{align*}$
But I'm afraid $$\left(\sum_{r\in\mathbb{F}_p}\widehat{f}(r)\left(1-\widehat{h}(r)\right)\right)^2\geq p^2\left(\sum_{r\in\mathbb{F}_p}\widehat{f}(r)^4\widehat{f}(-2r)^2\right)\cdot\left(\sum_{r\in\mathbb{F}_p}\left(1-\widehat{h}(r)^2\widehat{h}(-2r)\right)^2\right)$$ is not correct.
The inequality that you state is not what the authors claim (or require) -- what they write is $\lvert \Lambda(f)-\Lambda(g)\rvert\ll \sup_r\lvert \widehat{f}(r)-\widehat{g}(r)\rvert$. Here $\ll$ is the Vinogradov notation, where $f\ll g$ means $\lvert f\rvert\leq C\lvert g\rvert$ for some unspecified but absolute constant $C>0$.
Here's a proof that $\lvert \Lambda(f)-\Lambda(g)\rvert\leq 3\sup_r\lvert \widehat{f}(r)-\widehat{g}(r)\rvert,$ along the lines of what the authors intended.
Perhaps the easiest way to arrive at this kind of inequality is to think of $f$ as $g+(f-g)$, and to consider the more general expression
$$\Lambda(h_1,h_2,h_3)=\sum_r \widehat{h_1}(r)\widehat{h_2}(r)\widehat{h_3}(-2r).$$
This is a multilinear expression, and so it is natural to try expanding $\Lambda(f)$ successively using $f=g+(f-g)$, so that
$$\Lambda(f)= \Lambda(f,f,f)=\Lambda(f,f,g)+\Lambda(f,f,f-g)$$ $$=\Lambda(f,g,g)+\Lambda(f,f-g,g)+\Lambda(f,f,f-g)$$ $$=\Lambda(g)+\Lambda(f-g,g,g)+\Lambda(f,f-g,g)+\Lambda(f,f,f-g).$$
The final three summands here can all be bounded above by $\sup_r \lvert \widehat{f}(r)-\widehat{g}(r)\rvert$ using a combination of the triangle inequality, Parseval's identity,and (for $\Lambda(f,f-g,g)$) the Cauchy-Schwarz inequality.
For example, $$ \lvert\Lambda(f,f,f-g)\rvert\leq \sup_r \lvert \widehat{f}(r)-\widehat{g}(r)\rvert \sum_r \lvert \widehat{f}(r)\rvert^2\leq \sup_r \lvert \widehat{f}(r)-\widehat{g}(r)\rvert\frac{1}{p}\sum_x f(x)^2. $$ But since $f:\mathbb{F}_p\to \{0,1\}$, this right-hand side is at most $\sup_r \lvert \widehat{f}(r)-\widehat{g}(r)\rvert$.