Let $F$ be a finite field, 


For every $c \in F$, let $X\_1, X\_2,..., X\_9, Y\_1,..., Y\_9$ be independent non-zero random variables over $F$. 


Denote $X=(X\_1,...,X\_9)$, $Y=(Y\_1,...,Y\_9)$, also let $\langle X,Y \rangle =\sum\_{i}X\_i Y\_i$.




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**Question**: Show that
$$\sum\_{x,y: \langle x,y\rangle =c} (P[(X,Y)=(x,y) ])^{17/18} \leq 1.$$




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Remarks:


-feel free to swap $17/18$ for any other positive constant smaller then $1$.


-I can prove this for flat distributions (using min-entropy).


-I can prove it for dimension $2$ (with constant $1/2$ instead of $17/18$), that is for $X\_1, Y\_1$ instead of $18$ random variables.


I'd be super grateful for the proof or sketch or idea that actually works :-).


Best regards,


Maciej



Inner product over finite fields
This follows from [Young's inequality for convolutions](https://en.wikipedia.org/wiki/Young%27s_inequality#Young.27s_inequality_for_convolutions) which claims that $\|f\*g\|\_r\leq \|f\|\_p\|g\|\_q$ whenever $+\infty\geq p,q,r\geq 1$ and $\frac1p+\frac1q=\frac1r+1$. It can be generalized by the induction as
$$
 \|f\_1\*\dots\*f\_n\|\_r\leq \prod\_{i=1}^n \|f\_i\|\_{p\_i},
$$
whenever $\sum\_{i=1}^n\frac1{p\_i}=\frac1r+(n-1)$ and $+\infty\geq p\_1,p\_2,\dots,p\_n,r\geq 1$.


We regard $X\_i$, $Y\_i$ as functions $f\colon F\to[0,1]$ such that $\sum\_{a\in F} f(a)=1$. For such functions, we may define two different convolutions:
$$
 (f\*\_\times g)(a)=\sum\_{xy=a}f(x)g(y) \quad \text{and} \quad (f\*\_+ g)(a)=\sum\_{x+y=a}f(x)g(y).
$$
The first one works well on the functions vanishing at zero (and produces also a function vanishing at zero); the second one works well on all functions.


1. For every $i$ we have
$$
 1=\|X\_i^\alpha\|\_{1/\alpha}\|\cdot \|Y\_i^\alpha\|\_{1/\alpha}\geq \bigl\|X\_i^\alpha\*\_\times Y\_i^\alpha\bigr\|\_{1/(2\alpha-1)},
$$
as soon as $1/2\leq \alpha\leq 1$.
2. Next, we have
$$
 1\geq \prod\_{i=1}^n\bigl\|X\_i^\alpha\*\_\times Y\_i^\alpha\bigr\|\_{1/(2\alpha-1)}\\
 \geq \bigl\|(X\_1^\alpha\*\_\times Y\_1^\alpha)\*\_+\dots\*\_+(X\_n^\alpha\*\_\times Y\_n^\alpha)\bigr\|\_{1/(2n\alpha-(2n-1))},
$$
whenever $\frac{2n-1}{2n}\leq \alpha\leq 1$.


Finally, substitute $\alpha=\frac{2n-1}{2n}$. We get exactly what we need, and even a bit more:
$$
 1\geq \max\_{a\in F}\sum\_{x,y\in F^n\colon \langle x,y\rangle=a}\prod\_{i=1}^n X\_i(x\_i)^\alpha Y\_i(y\_i)^\alpha.
$$


