09-dec-moment-curve.tex

\section{Decoupling for the moment curve}

\subsection{Vinogradov mean value}
It is of interest in number theory to estimate Weyl sums such as
\[
\sum_{x=1}^{X} e(\xi_{1} x^{1} + \dotsb + \xi_{k}x^{k} ).
\]
Vinogradov's method deduces bounds for pointwise values (for fixed $\vec\xi = (\xi_{1},\dotsc,\xi_{k})$) from bounds for averages (\emph{mean values}) over all $\vec\xi$.
Specifically, it needs estimates on the moments
\begin{equation}
\label{eq:Vinogradov-mean-value}
\int_{\vec\xi \in [0,1]^{k}} \abs[\Big]{ \sum_{x=1}^{X} e(\xi_{1} x^{1} + \dotsb + \xi_{k}x^{k} ) }^{p} \dif\vec\xi.
\end{equation}
When the exponent $p=2s$ is an even integer, the expression \eqref{eq:Vinogradov-mean-value} can be interpreted as counting the solutions of a system of equations.
Indeed, the $2s$-power of the absolute value can be expanded, and each summand contributes $1$ to the integral if its frequency vanishes and $0$ otherwise.
The frequency vanishes iff the following system of equations holds.
\begin{equation}
\label{eq:Vinogradov-system}
\begin{split}
x_{1} + \dotsb + x_{s} &= x_{s+1} + \dotsb + x_{2s} \\
& \vdots \\
x_{1}^{k} + \dotsb + x_{s}^{k} &= x_{s+1}^{k} + \dotsb + x_{2s}^{k}.
\end{split}
\end{equation}
The number $J_{s,k}(X)$ of integer solutions to the system of equations \eqref{eq:Vinogradov-system} with all entries in the interval $[1,X]$ is certainly at least $X^{s}$ (considering the ``diagonal'' solutions $x_{s+j}=x_{j}$).
Another estimate comes from a more elaborate counting argument.
For $Y=(Y_{1},\dotsc,Y_{k})$ let $J_{s,k}(X,Y)$ be the number of integer solutions with all entries in $[1,X]$ of the system of equations
\begin{align*}
x_{1} + \dotsb + x_{s} &= Y_{1} \\
& \vdots \\
x_{1}^{k} + \dotsb + x_{s}^{k} &= Y_{k}.
\end{align*}
Then
\[
X^{s} = \sum_{Y_{1}=0}^{sX} \dots \sum_{Y_{k}=0}^{sX^{k}} J_{s,k}(X,Y).
\]
On the other hand,
\[
J_{s,k}(X) = \sum_{Y_{1}=0}^{sX} \dots \sum_{Y_{k}=0}^{sX^{k}} \bigl( J_{s,k}(X,Y) \bigr)^{2}.
\]
By the Cauchy--Schwarz inequality it follows that
\begin{multline*}
X^{s}
=
\sum_{Y_{1}=0}^{sX} \dots \sum_{Y_{k}=0}^{sX^{k}} J_{s,k}(X,Y)
\leq
\Bigl( \sum_{Y_{1}=0}^{sX} \dots \sum_{Y_{k}=0}^{sX^{k}} J_{s,k}(X,Y)^{2} \Bigr)^{1/2}
\Bigl( \sum_{Y_{1}=0}^{sX} \dots \sum_{Y_{k}=0}^{sX^{k}} 1 \Bigr)^{1/2}
\\ =
J_{s,k}(X)^{1/2} s^{k/2} X^{(1+\dotsb+k)/2}.
\end{multline*}
Combining this with the estimate for the number of diagonal solutions we obtain
\[
J_{s,k}(X) = \eqref{eq:Vinogradov-mean-value}
\geq \max (X^{s}, s^{-k} X^{2s - \frac{k(k+1)}{2}}).
\]
It turns out that this lower bound is essentially sharp, as proved in \cite{MR3548534} and \cite{MR3938716}.
We will present the decoupling proof from \cite{MR3548534} with some simplifications coming from later works \cite{MR3709122,MR3994585,MR4031117,arxiv:1902.03450}.
The two lower bounds that we obtained coincide when $p=2s=k(k+1)$, so this ``critical'' exponent can be expected to play a special role.
We will indeed consider only $p=k(k+1)$, since sharp results for other $p$'s can be obtained by interpolation.
\begin{remark}
The reduction to a unique critical exponent is special to the one-dimensional situation, in higher dimensions there may be many critical exponents.
In fact it becomes preferable not to single out any exponents.
\end{remark}


\subsection{Statement of the main result}
We will formulate a decoupling theorem for functions with Fourier support close to the unit moment curve $\Set{(\xi,\xi^{2},\dotsc,\xi^{k}) \given \xi \in [0,1]}$.
There is only one parameter, so cubes in the decoupling theorem for the paraboloid become just intervals.
We continue to denote by $\Part[Q]{\delta}$ the partition of an interval $Q$ into dyadic intervals of length $\delta$.
We omit $Q$ if $Q=[0,1]$.
We denote by $f_{\theta}$ a function whose Fourier support is adapted to the image of $\theta$ in the moment curve (this will be made precise when we describe the affine scaling procedure).
We denote by $\Dec_{k}(\delta)$ the smallest constant in the inequality
\begin{equation}
\label{eq:Dec-const-moment}
\norm{ \sum_{\theta \in \Part{\delta}} f_{\theta} }_{L^{k(k+1)}(\R^{k})} \leq \Dec_{k}(\delta) \ell^{2}_{\theta \in \Part{\delta}} \norm{ f_{\theta} }_{L^{k(k+1)}(\R^{k})}.
\end{equation}
We do not include the exponents $k(k+1)$ and $2$ in the notation since these will be the only exponents that we consider.
\begin{theorem}
\label{thm:Dec-moment}
For every $k\geq 1$ and $\epsilon>0$ we have $\Dec_{k}(\delta) \lesssim_{\epsilon} \delta^{-\epsilon}$.
\end{theorem}
Theorem~\ref{thm:Dec-moment} will be proved by induction on $k$.
The case $k=1$ is just $L^{2}$ orthogonality (in fact we have also already proved the case $k=2$, since this is the case of the one-dimensional paraboloid).
We will henceforth assume that $k\geq 2$ and Theorem~\ref{thm:Dec-moment} is already known for smaller values of $k$.

The main technical difficulties in the case of the moment curve are different from the case of the paraboloid.
The Bourgain--Guth argument is much easier, since all lower-dimensional contributions are zero-dimensional.
On the other hand, the treatment of multilinear terms becomes more sophisticated.

\subsubsection{Consequences for Vinogradov mean values}
By the usual procedure the inequality \eqref{eq:Dec-const-moment} can be localized to balls of radius $\gtrsim \delta^{-k}$.
Then we choose $\delta \sim X^{-1}$ and let each $\widehat{f_{\theta}}$ be supported in a point on the moment curve over a rational number with denominator $X$.
The right-hand side can then be easily computed since $\abs{f_{\theta}} \equiv \const$.
On the other hand, by periodicity the left-hand side coincides up to scaling with \eqref{eq:Vinogradov-mean-value}.
After scaling this gives the estimate
\[
\int_{\vec\xi \in [0,1]^{k}} \abs[\Big]{ \sum_{x=1}^{X} e(\xi_{1} x^{1} + \dotsb + \xi_{k}x^{k} ) }^{k(k+1)} \dif\vec\xi
\lesssim_{\epsilon}
X^{k(k+1)/2+\epsilon}.
\]

\subsubsection{Affine scaling}
\label{sec:scaling}
Let $\theta \in \Part[\sigma]$ with left endpoint $c=c(\theta)$.
Consider the affine transformation
\begin{equation}
\label{eq:affine-scaling:space}
(L_{\theta}(x))_{j} =
\sigma^{j} \sum_{i : 0 \leq i \leq j} \binom{i}{j} (c)^{j-i} x_{i},
\quad
1 \leq j \leq k,
\end{equation}
where we set $x_{0}=1$.
This transformation preserves the moment curve and maps the point $0$ to the image of $c$ in the moment curve.
The support condition that we impose on our functions is that $\supp \widehat{f_{\theta}}$ is contained in the image of a fixed ball centered at the origin under $L_{\theta}$.
Then for $\theta_{0} \in \Part{\delta_{0}}$ we immediately obtain the rescaled decoupling inequality
\[
\norm{ \sum_{\theta' \in \Part[\theta_{0}]{\delta_{0}\delta_{1}}} f_{\theta'} }_{L^{k(k+1)}(\R^{k})}
\leq
\Dec_{k}(\delta_{1}) \ell^{2}_{\theta' \in \Part[\theta]{\delta_{0}\delta_{1}}} \norm{ f_{\theta'} }_{L^{k(k+1)}(\R^{k})}.
\]

\subsection{Transversality}
\label{sec:transversality}
The functions $f_{\theta}$ have Fourier support in boxes of size $\delta \times \delta^{2} \times \dotsm \times \delta^{k}$.
Hence they are morally constant on boxes of size $\delta^{-1} \times \delta^{-2} \times \dotsm \times \delta^{-k}$.
We need a description of orientation of these boxes.
To this end we will use the higher order tangent spaces
More precisely, we use the $l$-th order tangent spaces
\begin{equation}\label{order_tangent}
V^{(l)}(t):=\lin \Set{\partial^{j} \Phi(t) \given 1 \leq j \leq l} \subseteq \R^{k},
\quad t\in [0, 1],
\end{equation}
where $\Phi(t) = (t^{\gamma})_{1\leq \gamma\leq k}$ parametrizes the moment curve.
If $t\in\theta$, then the function $f_{\theta}$ is morally constant at scale $\delta^{-l-1}$ in the orthogonal direction to $V^{(l)}(t)$.

In order to make use of Kakeya--Brascamp--Lieb inequalities we have to verify that the spaces $V^{(l)}(t)$ are transverse when we consider sufficiently widely spaced $t$'s.
Due to the lack of an explicit description of BL constants, transversality in this case means that the BCCT condition for finiteness of BL constants is satisfied.
In contrast to the paraboloid case it is not a priori clear how many different $t$'s one would have to consider to achieve such transversality.

\subsubsection{Projections onto higher order tangential spaces}
In the case of the moment curve we get lucky and it turns out that for any subspace $V \subseteq \R^{k}$ the projection onto $V^{(l)}(t)$ almost always has maximal possible dimension.
\begin{theorem}
\label{thm:moment-rank}
For each $k\ge 1$ there exists $M_{0,k}$ such that for every subspace $V \subseteq \R^{k}$ and every $1\leq l\leq k$ we have
\begin{equation}
\label{eq:moment-rank}
\dim \pi_{V^{(l)}(t)} V = \min(l,\dim V)
\end{equation}
for all but at most $M_{0,k}$ values of $t \in [0,1]$.
\end{theorem}
\begin{proof}
Decreasing the dimension of $V$ or $l$ if necessary we may assume without loss of generality $\dim V = l$.
Fix a basis $(v_{1},\dotsc,v_{l})$ of $V$ such that $a_{h} := \max \Set{ j \given v_{h,j} \neq 0 }$ is strictly decreasing in $h=1,\dotsc,l$.
The dimension on LHS\eqref{eq:moment-rank} equals the rank of the $l\times l$ matrix
\begin{equation}
\calM_V^{(l)}(t)
:=
\bigl( v_1, \dotsc, v_{l} \bigr)^T \times \bigl( \partial^{j} \Phi(t) \bigr)_{1 \leq j \leq l}
\end{equation}
over $\R$.
We claim that the determinant of this matrix is a non-zero polynomial in $t$.
This will suffice to establish the claim since the degree of this polynomial is bounded by some $M_{0,k}$, so it will have at most $M_{0,k}$ zeros.
If $t$ is not a zero of this poynomial, then $\calM_{V}^{(l)}(t)$ has real rank $l$.

To simplify notation write $f_{v}(x) := \sum_{i=1}^{l} v_{i}x^{i}$.
Then
\[
\calM_V^{(l)}(t) = ( \partial^{j} f_{v_h} )_{1 \leq j,h \leq l}.
\]
Note that $\deg f_{h} = a_{h}$ is striclty decreasing in $h$.
Applying column and row transformations over the field of rational functions $\R(x)$ we obtain the matrix
\[
( x^{-a_{h}} x^{j} \partial^{j} f_{v_h} )_{1 \leq j,h \leq l}.
\]
All entries of this matrix are linear combinations of non-positive powers of $x$.
Hence the constant term in its determinant equals the determinant of constant terms, which is given by
\[
\det ( (a_{h} \dotsm (a_{h}-j+1)) v_{h,a_{h}} )_{1 \leq j,h \leq l}.
\]
Since all $v_{h,a_{h}}$ are non-zero this determinant is non-zero iff
\[
\det ( a_{h} \dotsm (a_{h}-j+1) )_{1 \leq j,h \leq l}
\]
is non-zero.
In the $j$-th row all entries are polynomials in $a_{h}$ of degree $j$.
By row operations one can bring this determinant in the form
\[
\det ( a_{h}^{j} )_{1 \leq j,h \leq l}.
\]
But this is $a_{1}\dotsm a_{h}$ times a Vandermonde determinant, so non-zero.
\end{proof}



\subsubsection{Verification of BCCT condition}
\begin{corollary}
For every $k$ there exists $M$ such that for any distinct $t_{1},\dotsc,t_{M} \in [0,1]$ and any $1 \leq l < k$ we have
\[
BL((V^{(l)}(t_{j}))_{j=1}^{M}) < \infty
\]
and this BL datum is simple.
\end{corollary}
It is not necessary to ensure simplicity to proceed with the proof, but we only proved the BCCT criterion for finiteness of BL constants in the simple case.
\begin{proof}
By the BCCT condition the BL constant is finite if for every subspace $V \subseteq \R^{k}$ we have
\begin{equation}
\tag{BCCT}
\dim V \leq \frac{k}{l M} \sum_{m=1}^{M} \dim \pi_{V^{(l)}(t_{j})} V
\end{equation}
with equality for $V=\R^{k}$.
We have actually only proved this in the simple case when the inequality is strict for $0 < \dim V < k$, and we will be able to put ourselves in this situation.
The equality in the case $\dim V = k$ is easy to see.

Assume now $\dim V < k$.
By Theorem~\ref{thm:moment-rank} we have
\[
\dim \pi_{V^{(l)}(t_{j})} V = \min(l, \dim V)
\]
for all but at most $M_{0,k}$ many $t$'s.
Hence
\begin{multline*}
RHS(BCCT) \geq
\frac{k (M-M_{0,k})}{l M} \min(l, \dim V)
=
\frac{M-M_{0,k}}{M} \min(k, \frac{k}{l} \dim V)\\
\geq
\frac{M-M_{0,k}}{M} \frac{k}{k-1} \dim V,
\end{multline*}
where we used $\dim V \leq k-1$ in the last step.
So it suffices to choose $M$ large enough so that $\frac{M-M_{0,k}}{M} \frac{k}{k-1} > 1$.
\end{proof}

\begin{corollary}
\label{cor:not-clustered-implies-transverse}
For every $K$ there exists $\nu=\nu_{K}$ such that any $K^{-1}$-separated $\alpha_1, \dotsc, \alpha_M \in \Part{K^{-1}}$ are \emph{$\nu$-transverse} in the sense that for every $1 \leq l < k$ and any $x_{j} \in \alpha_{j}$ we have
\[
\BL((V^{(l)}(x_{j}))_{j=1}^{M})
\leq \nu^{-1}.
\]
\end{corollary}
\begin{proof}
We have already seen that the BL constants are finite.
The uniform upper bound follows from the fact that the BL constants are locally bounded on the set where they are finite and compactness.
\end{proof}

\begin{remark}
The above compactness argument is ineffective.
It would be desirable to replace it by an explicit estimate for BL constants.
\end{remark}

\subsection{Bourgain--Guth argument}
We will work with $M$-linear expressions with $M$ given by Corollary~\ref{cor:not-clustered-implies-transverse}.
We denote $\avprod A_{i} := \avprod_{i} A_{i} := \prod_{i=1}^{M} A_{i}^{1/M}$ and $p:=k(k+1)$.

For a positive integer $K$ and $0 < \delta < K^{-1}$ we denote by $\MulDec_{k}(\delta, K)$ the smallest constant such that the inequality
\begin{equation}
\label{eq:multilin-dec-const-KM}
L^{p}_{x\in \R^{k}} \avprod \norm{ f_{\alpha_{i}} }_{\avL^{p}(B(x,K))}\\
\le \MulDec_{k}(\delta, K)
\avprod \ell^{2}_{\theta \in \Part[\alpha_i]{\delta}} \norm{ f_{\theta} }_{L^p(\R^{k})}
\end{equation}
holds for every $\nu_{K}$-transverse tuple $\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}}$.

\begin{theorem}
\label{thm:multilinear-to-linear}
For each $\epsilon>0$ there exists $K\geq 1$ such that for all $0 < \delta < 1$ we have
\begin{equation}
\Dec(\delta)
\lesssim
\delta^{-\epsilon}
+ \delta^{-\epsilon} \max_{\delta\le \delta'\le 1}(\delta/\delta')^{-\epsilon} \MulDec(\delta', K).
\end{equation}
\end{theorem}
Theorem~\ref{thm:multilinear-to-linear} is obtained by iterating Corollary~\ref{cor:bourgain-guth-arg:scaled} that is a rescaled version of the following Proposition~\ref{prop:bourgain-guth-arg}.
This iteration goes back to \cite{MR2860188}.
\begin{proposition}
\label{prop:bourgain-guth-arg}
For every $0<\delta<K^{-1}$ we have
\begin{equation}
\label{eq:BG-arg}
\norm{f}_{p}
\lesssim
\ell^{2}_{\alpha\in \Part{K^{-1}}} \norm{ f_{\alpha} }_{p}
+ K^{M+1} \MulDec(\delta, K) \ell^{2}_{\theta \in \Part{\delta}} \norm{ f_{\theta} }_{p}
\end{equation}
\end{proposition}
\begin{proof}[Proof of Proposition~\ref{prop:bourgain-guth-arg}]
Let $B \subset \R^{k}$ be a ball of radius $K$ and
\[
S_{B} := \ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{\avL^{p}(B)}.
\]
If $\norm{ f }_{\avL^{p}(B)} \geq 2M S_{B}$, then there exist at least $2M$ cubes $\alpha \in \Part{K^{-1}}$ with
\[
\norm{ f_\alpha }_{\avL^{p}(B')} \geq (2K)^{-1} \norm{ f }_{\avL^{p}(B)}.
\]
So we can choose $M$ cubes $\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}}$ that are $K^{-1}$-separated and
\[
\norm{ f }_{\avL^{p}(B)} \leq K \avprod \norm{ f_{\alpha_{i}} }_{L^{p}(B)}.
\]
Hence in any case
\[
\norm{ f }_{\avL^{p}(B)} \lesssim
\ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{\avL^{p}(B)}
+
K \sum_{\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}}} \avprod \norm{ f_{\alpha_{i}} }_{L^{p}(B)},
\]
where the sum runs over all $K^{-1}$-separated tuples.
Integrating this inequality over $B$ we obtain
\begin{align*}
\norm{ f }_{L^{p}(\R^{k})}
&=
L^{p}_{x \in \R^{k}} \norm{ f }_{\avL^{p}(B(x,K))}
\\ &\leq
C L^{p}_{x \in \R^{k}} \ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{\avL^{p}(B(x,K))}
+
K L^{p}_{x \in \R^{k}} \sum_{\substack{\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}} \\ K^{-1}\text{-separated}}} \avprod \norm{ f_{\alpha_{i}} }_{L^{p}(B(x,K))}
\\ &\leq
C \ell^{2}_{\alpha\in\Part{K^{-1}}} L^{p}_{x \in \R^{k}} \norm{ f_\alpha }_{\avL^{p}(B(x,K))}
+
K \sum_{\substack{\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}} \\ K^{-1}\text{-separated}}} L^{p}_{x \in \R^{k}} \avprod \norm{ f_{\alpha_{i}} }_{L^{p}(B(x,K))}
\\ &\leq
C \ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{L^{p}(\R^{k})}
+
K \sum_{\substack{\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}} \\ K^{-1}\text{-separated}}} \Dec_{k}(\delta,K) \avprod \ell^{2}_{\theta \in \Part[\alpha_{i}]{\delta}} \norm{ f_{\theta} }_{L^{p}(\R^{k})}
\\ &\leq
C \ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{L^{p}(\R^{k})}
+
K \sum_{\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}}} \Dec_{k}(\delta,K) \avprod \ell^{2}_{\theta \in \Part{\delta}} \norm{ f_{\theta} }_{L^{p}(\R^{k})}
\\ &\leq
C \ell^{2}_{\alpha\in\Part{K^{-1}}} \norm{ f_\alpha }_{L^{p}(\R^{k})}
+
K^{M+1} \Dec_{k}(\delta,K) \ell^{2}_{\theta \in \Part{\delta}} \norm{ f_{\theta} }_{L^{p}(\R^{k})}.
\qedhere
\end{align*}
\end{proof}

\begin{corollary}
\label{cor:bourgain-guth-arg:scaled}
For $0 < \delta < K^{-1}$ we have
\begin{equation}
\label{eq:BG-arg:scaled}
\Dec(\delta)
\leq \max\Bigl( C \Dec(K \delta), K^{M+1} \MulDec(\delta, K) \Bigr).
\end{equation}
\end{corollary}

\begin{proof}[Proof of Theorem~\ref{thm:multilinear-to-linear}]
Choose $K \in 2^{k\N}$ so large that the $C$ on the right-hand side of \eqref{eq:BG-arg:scaled} is bounded by $K^{\epsilon}$.
For $\delta < 1/K$ iterate the inequality \eqref{eq:BG-arg:scaled} $\floor[\big]{\frac{\log \delta}{\log K}}$ times and use a trivial estimate for $\Dec$ at the end.
\end{proof}

From Theorem~\ref{thm:multilinear-to-linear} it follows that if for some $\eta \geq 0$, all $K\in 2^{\N}$, and all $0 < \delta < 1$ we have
\begin{equation}
\label{eq:multlin-dec-power}
\MulDec(\delta, K)
\lesssim_{K}
\delta^{-\eta},
\end{equation}
then we obtain
\begin{equation}
\label{eq:lin-dec-power-iteration}
\Dec(\delta)
\lesssim_{\epsilon}
\delta^{-\eta - \epsilon}
\end{equation}
for every $\epsilon>0$.
Let $\eta$ be the smallest exponent in the decoupling inequality.

\subsection{Induction on scales}
\label{sec:induction-on-scales}
We fix $K^{-1}$-separated intervals $\alpha_{1},\dotsc,\alpha_{M} \in \Part{K^{-1}}$ and functions $f_{\theta}$.
We write
\[
n_{l} = l,
\quad
\calK_{l} = 1+\dotsb+l = \frac{l(l+1)}{2},
\quad
p = k(k+1).
\]
For $1 \leq l \leq k$ define
\begin{align*}
p_{l} &:= p \frac{\calK_{l}}{\calK_{k}} = l(l+1),\\
t_{l} &:= p \frac{n_{l}}{n_{k}} = l(k+1).
\end{align*}
Here $p_{l}$ is the sharp decoupling exponent for the $l$-th moment curve and $t_{l}$ is an exponent in the BL inequality that we will use.

Define $\alpha_l$ and $\beta_l$ by
\begin{align}
\label{eq:alpha}
\frac{1}{\frac{n_{l}}{n_{k}}}
&=
\frac{\alpha_l}{\frac{n_{l+1}}{n_{k}}}+\frac{1-\alpha_l}{\frac{\calK_{l}}{\calK_{k}}},
& 1 \leq l < k,\\
\label{eq:beta}
\frac{1}{\frac{\calK_{l}}{\calK_{k}}}
&=
\frac{1-\beta_l}{\frac{\calK_{l-1}}{\calK_{k}}}+\frac{\beta_l}{\frac{n_{l}}{n_{k}}},
& 1 < l < k,
\end{align}
and $\beta_{1}:=1$.

For $2 \leq t \leq p$, $0<b<1$, and $b<s$ let
\begin{equation}
\label{eq:A}
A_{t} (b, s)
:=
L^{p}_{x} \avprod \ell^{2}_{\beta \in \Part[\alpha_i]{\delta^{b}}} \norm{f_{\beta}}_{\avL^t(w_{B(x,\delta^{-s})})}.
\end{equation}
The induction on scales argument will involve the quantities
\[
A_{t(l)}(b) := A_{t_{l}}(b,lb),
\]
\[
A_{p(l)}(b) := A_{p_{l}}(b,(l+1)b).
\]
Here $t(l)$ and $p(l)$ are formal expressions and can be read ``of type $t$ with degree $l$'' and ``of type $p$ with degree $l$''.
For $0<b<1$ and $*=t(l),p(l)$ let
\[
a_{*}(b) := \inf \Set{ a \given A_{*}(b) \lesssim_{a,K} \delta^{-a} RHS\eqref{eq:multilin-dec-const-KM} \text{ for all } K }.
\]

\subsubsection{Linear decoupling}
We can use H\"older to eliminate all multilinearity and use the linear decoupling estimate.
For $1 \leq t \leq p$ and $1 \leq l \leq k$ this gives the bound
\begin{equation}
\label{eq:A<prod}
\begin{split}
A_{t}(b,s)
&=
L^{p}_{x} \avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta^{b}}} \norm{ f_{\beta} }_{\avL^{t}(w_{B(x,\delta^{-lb})})}
\\ &\leq
\avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta^{b}}} L^{p}_{x} \norm{ f_{\beta} }_{\avL^{p}(w_{B(x,\delta^{-lb})})}
\\ &=
\avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta^{b}}} \norm{ f_{\beta} }_{L^{p}(\R^{k})}
\\ &\leq
\Dec(\delta^{1-b})
\avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta}} \norm{ f_{\beta} }_{p}
\\ &\lesssim_{\epsilon}
\delta^{-\eta(1-b)-\epsilon}
\avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta}} \norm{ f_{\beta} }_{p}.
\end{split}
\end{equation}
This shows
\begin{equation}
\label{eq:a*:linear-dec}
a_{*}(b) \leq \eta(1-b).
\end{equation}

\subsubsection{Bourgain--Guth argument}
First we estimate the left-hand side of \eqref{eq:multilin-dec-const-KM} by the quantities involved in the iterative procedure.
For $1 \leq l \leq k$, $1 \leq t \leq p$, $0<b \leq s$, and $\delta$ sufficiently small so that $\delta^{-s}\geq K$ we have
\begin{equation}
\label{eq:multlin<A}
\begin{split}
LHS\eqref{eq:multilin-dec-const-KM}
&=
L^{p}_{x\in \R^{k}} \avprod \norm{ f_{\alpha_{i}} }_{\avL^{p}(B(x,K))}
\\ &\lesssim
L^{p}_{x\in \R^{k}} \avprod \norm{ f_{\alpha_{i}} }_{\avL^{p}(B(x,\delta^{-s}))}
\\ &\leq
L^{p}_{x\in \R^{k}} \avprod \sum_{\beta \in \Part[\alpha_{i}]{\delta^{b}}} \norm{ f_{\beta} }_{\avL^{p}(B(x,\delta^{-s}))}
\\ &\lesssim
\delta^{-b/2-(s-b)k(1/t-1/p)} L^{p}_{x\in \R^{k}} \avprod \ell^{2}_{\beta \in \Part[\alpha_{i}]{\delta^{b}}} \norm{ f_{\beta} }_{\avL^{t}(w_{B(x,\delta^{-s})})}
\\ &\leq
\delta^{-Cb} A_{*}(b).
\end{split}
\end{equation}
Here we have used the reverse H\"older inequality (Corollary~\ref{cor:rev-holder}) to estimate the $\avL^{p}$ norm by the $\avL^{t}$ with some loss.
This shows
\begin{equation}
\label{eq:a*:BG}
\eta \leq Cb + a_{*}(b).
\end{equation}

\subsubsection{Ball inflation}
Similar to the paraboloid case we obtain the following results from Kakeya--Brascamp--Lieb inequalities.

\begin{lemma}[Ball inflation]
\label{lem:ball-inflation}
Let $1\le l < k$, $1 \leq t < \infty$.
Let $\rho \leq 1/K$ and let $B \subset \R^{k}$ be a ball of radius $\rho^{-(l+1)}$.
Then we have
\begin{equation}
\label{eq:ball-inflation}
\avL^{t \frac{k}{l}}_{x\in B} \avprod \ell^{t}_{\beta \in \Part[\alpha_i]{ \rho}} \norm{f_{\beta}}_{\avL^{t}(w_{B(x,\rho^{-l})})}
\lesssim_{\nu,\epsilon} \rho^{-\epsilon}
\avprod \ell^{t}_{\beta \in \Part[\alpha_i]{\rho}} \norm{ f_{\beta} }_{\avL^{t}(w_B)}.
\end{equation}
\end{lemma}

\begin{corollary}[Ball inflation]
\label{cor:ball-inflation}
Let $1\le l < k$, $1 \leq q \leq t < \infty$.
Let $\rho \leq 1/K$ and let $B \subset \R^{k}$ be a ball of radius $\rho^{-(l+1)}$.
Then we have
\begin{equation}
\label{eq:ball-inflation}
\avL^{t \frac{k}{l}}_{x\in B} \avprod \ell^{q}_{\beta \in \Part[\alpha_i]{ \rho}} \norm{f_{\beta}}_{\avL^{t}(w_{B(x,\rho^{-l})})}
\lesssim_{\nu,\epsilon} \rho^{-\epsilon}
\avprod \ell^{q}_{\beta \in \Part[\alpha_i]{\rho}} \norm{ f_{\beta} }_{\avL^{t}(w_B)}.
\end{equation}
\end{corollary}

Note that by \eqref{eq:alpha} for $1\leq l < k$ we have
\begin{equation}
\label{eq:alpha-ineq}
\frac{1}{t_{l}}
=
\frac{\alpha_l}{t_{l+1}}+\frac{1-\alpha_l}{p_{l}}.
\end{equation}


For $1 \leq l < k$ by Corollary~\ref{cor:ball-inflation} and H\"older's inequality together with \eqref{eq:alpha-ineq} we obtain
\begin{equation}
\label{eq:est1}
\begin{split}
A_{t(l)}(b)
&=
A_{t_{l}}(b,lb)
\\ &\lesssim_{\epsilon,K}
\delta^{-b\epsilon} A_{t_{l}}(b,(l+1)b)
\\ &\lesssim
\delta^{-b\epsilon} A_{ t_{l+1}}(b,(l+1)b)^{\alpha_{l}}
A_{ p_{l}}(b,(l+1)b)^{1-\alpha_{l}}
\\ &=
\delta^{-b\epsilon} A_{t(l+1)}(b)^{\alpha_{l}} A_{p(l)}(b)^{1-\alpha_{l}}.
\end{split}
\end{equation}
This implies
\begin{equation}
\label{eq:a*:ball-inflation}
a_{t(l)}(b) \leq \alpha_{l} a_{t(l+1)}(b) + (1-\alpha_{l}) a_{p(l)}(b).
\end{equation}

\subsubsection{Lower degree decoupling}
By~\eqref{eq:beta} for $1 < l < k$ we have
\begin{equation}
\label{eq:beta-ineq}
\frac{1}{p_{l}}
=
\frac{1-\beta_l}{p_{l-1}}+\frac{\beta_l}{t_{l}}.
\end{equation}

For $1 \leq l < k$ by the localized version of the decoupling inequality for the $l$-th moment curve and H\"older's inequality with \eqref{eq:beta-ineq} we obtain
\begin{equation}
\label{eq:est2}
\begin{split}
A_{p(l)}(b)
&=
A_{p_{l}}(b,(l+1)b)
\\ &=
L^{p}_{x} \avprod \ell^{2}_{\beta \in \Part[\alpha_i]{\delta^{b}}} \norm{f_{\beta}}_{\avL^{p_{l}}(w_{B(x,\delta^{-(l+1)b})})}
\\ &\lesssim_{\epsilon}
\delta^{-\epsilon b/l} L^{p}_{x} \avprod \ell^{2}_{\beta \in \Part[\alpha_i]{\delta^{(l+1)b/l}}} \norm{f_{\beta}}_{\avL^{p_{l}}(w_{B(x,\delta^{-(l+1)b})})}
\\ &\leq
\delta^{-\epsilon b/l}
A_{t(l)}(\frac{(l+1)b}{l})^{\beta_{l}} A_{p(l-1)}(\frac{(l+1)b}{l})^{1-\beta_{l}}.
\end{split}
\end{equation}
Note that this also holds for $l=1$ because $p_{1} \leq t_{1}$.
This implies
\begin{equation}
\label{eq:a*:lower-deg-dec}
a_{p(l)}(b) \leq \beta_{l} a_{t(l)}((l+1)b/l) + (1-\beta_{l}) a_{p(l-1)}((l+1)b/l)
\end{equation}
for $0<b<l/(l+1)$.

\subsubsection{Wrapping up the induction}
\begin{proposition}
The inequalities \eqref{eq:a*:linear-dec}, \eqref{eq:a*:BG}, \eqref{eq:a*:ball-inflation}, and \eqref{eq:a*:lower-deg-dec} imply $\eta \leq 0$.
\end{proposition}
\begin{proof}
We eliminate the dependence on $b$ by setting\footnote{This definition is from Tao's blog post\\\url{https://terrytao.wordpress.com/2019/06/14/abstracting-induction-on-scales-arguments/}}
\[
\tilde{a}_{*} := \liminf_{b \to 0} \frac{\eta - a_{*}(b)}{b}.
\]
The hypotheses then imply
\[
\tilde{a}_{*} \geq \eta.
\]
\[
\tilde{a}_{*} \leq C.
\]
\[
\tilde{a}_{t(l)} \geq \alpha_{l} \tilde{a}_{t(l+1)} + (1-\alpha_{l}) \tilde{a}_{p(l)}.
\]
\[
\tilde{a}_{p(l)}(b) \geq \frac{l+1}{l} \bigl( \beta_{l} \tilde{a}_{t(l)} + (1-\beta_{l}) \tilde{a}_{p(l-1)} \bigr).
\]
Using \eqref{eq:a*:lower-deg-dec} let us verify the last of these inequalities, which is also the least obvious:
\begin{align*}
\tilde{a}_{p(l)}(b)
&=
\liminf_{b\to 0} \frac{\eta - a_{p(l)}(b)}{b}
\\ &\geq
\liminf_{b\to 0} \frac{\eta - \beta_{l} a_{t(l)}((l+1)b/l) - (1-\beta_{l})a_{p(l-1)}((l+1)b/l)}{b}
\\ &\geq
\frac{l+1}{l} \beta_{l} \liminf_{b\to 0} \frac{\eta - a_{t(l)}((l+1)b/l)}{(l+1)b/l}
+
\frac{l+1}{l} (1-\beta_{l}) \liminf_{b\to 0} \frac{\eta - a_{p(l-1)}((l+1)b/l)}{(l+1)b/l}
\\ &=
\frac{l+1}{l} \bigl( \beta_{l} \tilde{a}_{t(l)} + (1-\beta_{l}) \tilde{a}_{p(l-1)} \bigr).
\end{align*}
Let
\[
W := \sum_{l=1}^{k-1} ( l \tilde{a}_{p(l)} + 2 l \tilde{a}_{t(l)} ).
\]
Applying all our inequalities we get
\begin{align*}
W & \geq
\sum_{l=1}^{k-1} (l+1) \bigl( \beta_{l} \tilde{a}_{t(l)} + (1-\beta_{l}) \tilde{a}_{p(l-1)} \bigr)
+
\sum_{l=1}^{k-1} 2 l (\alpha_{l} \tilde{a}_{t(l+1)} + (1-\alpha_{l}) \tilde{a}_{p(l)})
\\ &=
2(k-1)\alpha_{k-1} \tilde{a}_{t(k)}
+ \sum_{l=1}^{k-1} ( (l+1) \beta_{l} + 2 (l-1) \alpha_{l-1} ) \tilde{a}_{t(l)}
\\ & \quad + \sum_{l=1}^{k-1} ( (l+2)(1-\beta_{l+1}) + 2l (1-\alpha_{l})) \tilde{a}_{p(l)},
\end{align*}
where by convention $\alpha_{0}=0$, $\beta_{k}=1$.
Now from \eqref{eq:alpha} and \eqref{eq:beta} we compute
\[
\frac{1}{l/k}
=
\frac{\alpha_l}{(l+1)/k}+\frac{1-\alpha_l}{l(l+1)/(k(k+1))},
\quad
\frac{1}{l(l+1)/(k(k+1))}
=
\frac{1-\beta_l}{(l-1)l/(k(k+1))}+\frac{\beta_l}{l/k}.
\]
\[
(l+1)/(k+1)
=
\frac{l}{k+1}\alpha_l+(1-\alpha_l),
\quad
\frac{1}{(l+1)/(k+1)}
=
\frac{1-\beta_l}{(l-1)/(k+1)}+\beta_l.
\]
\[
\alpha_{l} = \frac{1-(l+1)/(k+1)}{1-l/(k+1)},
\quad
\beta_{l} = \frac{(k+1)/(l+1)-(k+1)/(l-1)}{1-(k+1)/(l-1)}
\]
\[
\alpha_{l} = \frac{k-l}{k+1-l},
\quad
\beta_{l} = \frac{(k+1)(l-1) - (l+1)(k+1)}{((l-1)-(k+1))(l+1)}
=
\frac{2(k+1)}{(l+1)(k-l+2)}
\]
For $1<l<k$ we get
\begin{multline*}
(l+1)\beta_{l} + 2(l-1)\alpha_{l-1}
=
\frac{2(k+1)}{k-l+2} + 2(l-1)\frac{k-l+1}{k-l+2}
\\ =
2l + \frac{2(k+1)}{k-l+2} -\frac{k-l+1}{k-l+2} - 2l\frac{1}{k-l+2}
=
2l.
\end{multline*}
For $l=1$ this is easier.
For $1\leq l \leq k-2$ we get
\begin{multline*}
(l+2)(1-\beta_{l+1}) + 2l (1-\alpha_{l})
=
(l+2)-\frac{2(k+1)}{k-l+1} + 2l \frac{1}{k-l+1}
=
l.
\end{multline*}
For $l=k-1$ this is again easier.
Hence we get
\[
W \geq W + 2(k-1)\alpha_{k-1} \tilde{a}_{t(k)},
\]
so $\eta \leq \tilde{a}_{t(k)} \leq 0$.
\end{proof}
\begin{remark}
The coefficients used to define $W$ are the Perron--Frobenius eigenvector of the matrix of coefficients of the inequalities for $\tilde{a}_{*}$'s.
\end{remark}