Inner product over finite fields

Question

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 $\langle X,Y \rangle =\sum_{i}X_i Y_i$.

---

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

---

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

Answer 1

This follows from Young's 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: $$ 1\geq \max_{a\in F}\sum_{x,y\in F^n\colon \langle x,y\rangle=a}\prod_{i=1}^n X_i(x_i)^\alpha Y_i(y_i)^\alpha. $$