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As for the second part, I think something like this works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C(p)\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to rewrite the left member of (3.4) as $$ \big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big] + R\big|, $$ where $R = \pm\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)\pm\dots$ is the remainder we get by algebraically manipulating this expression to match the given left-hand side of (3.4). Writing (3.4) this way sets us up to use the first part of the proof, and indeed the part not involving $R$ agrees with what Bourgain wrote. To see how to handle $R$, I'll just demonstrate what to do with $\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)$, since the triangle inequality finishes the bound. By Cauchy-Schwarz, \begin{align*} |\E[\phi_1(X)\phi_2(Y)]|\le (\E|\phi_1(X)|^2)^{\frac12}(\E|\phi_2(Y)|^2)^{\frac12}. \end{align*} By the bound (3.1), $|\phi_1(X)|\lesssim (1+|X|)^{p_1}\le(1+|X-\E X|)^{p_1}(1+|\E X|)^{p_1}$, so by Khinchin's inequality, $\E[|\phi_1(X)|^2]\lesssim_{p_1}|x|^{2p_1}(1+|\mathbb E X|)^{2p_1}\le(1+|\E X|)^{2p_1}$. A similar bound applies to $\E|\phi_2(Y)|^2$, of course. To estimate $|\phi_3(\E Z)|$, use (3.1) again to get $(1+|\E Z|)^{p_3}$, and use $p_1+p_2-\delta>0$ to get the final bound $|\phi_3(\E Z)|\lesssim (1+|\sum z_i|)^{p-\delta}$, which is smaller than the right-hand side of (3.4). (Note how the constant doesn't depend on $n$, the number of terms, because we used Khinchin's inequality.)

The last thing to check is $$\big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big]\big|\le \text{RHS (3.4)}.$$ For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately, then, using this inequality, the bound we get is \begin{align*} &(1+|\sum x_i| + |\sum y_i| + |\sum z_i|)^{p-\delta}\\ &\quad\times \int(1+|\sum(\eta_i^1-\frac13)x_i|+|\sum(\eta_i^2-\frac13)y_i|+|\sum(\eta_i^3-\frac13)z_i|)^p. \end{align*} Use Khinchin's inequality again to show the term on the second line of the above display is $\lesssim_p 1$, finishing the proof.

As for the second part, the following works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C(p)\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to use the identity $$ ABC - abc = (A-a)BC + a(B-b)C + ab(C - c) $$ with $A=\phi_1(X),B=\phi_2(Y),C=\phi_3(Z)$, and $a = \phi_1(\E X),b=\phi_2(\E Y),c=\phi_3(\E Z)$, and use the triangle inequality, estimating the three addends on the right-hand side using the assumptions on the $\phi_\alpha$. For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately then, the rest of the argument is using Khinchine's inequality and the conditions on the $\phi_\alpha$ to match the inequality (3.4).

For example, \begin{align*} \E(1+|X-\E X|)^p \le (1+C(p)|x|)^p < c, \end{align*} and so on.

As for the second part, I think something like this works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to rewrite the left member of (3.4) as $$ \big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big] + R\big|, $$ where $R = \pm\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)\pm\dots$ is the remainder we get by algebraically manipulating this expression to match the given left-hand side of (3.4). Writing (3.4) this way sets us up to use the first part of the proof, and indeed the part not involving $R$ agrees with what Bourgain wrote. To see how to handle $R$, I'll just demonstrate what to do with $\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)$, since the triangle inequality finishes the bound. By Cauchy-Schwarz, \begin{align*} |\E[\phi_1(X)\phi_2(Y)]|\le (\E|\phi_1(X)^2|)^{\frac12}(\E|\phi_2(Y)^2|)^{\frac12}. \end{align*} By the bound (3.1), $|\phi_1(X)|\lesssim (1+|X|)^{p_1}$, so by Khinchin's inequality, $\E|\phi_1(X)^2|\lesssim_{p_1}|x|^{2p_1}\le1$. A similar bound applies to $\E|\phi_2(Y)|^2$, of course. To estimate $|\phi_3(\E Z)|$, use (3.1) again to get $(1+|\E Z|)^{p_3}$, and use $p_1+p_2-\delta>0$ to get the final bound $|\phi_3(\E Z)|\lesssim (1+|\sum z_i|)^{p-\delta}$, which is smaller than the right-hand side of (3.4). (Note how the constant doesn't depend on $n$, the number of terms, because we used Khinchin's inequality.)

The last thing to check is $$\big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big]\big|\le \text{RHS (3.4)}.$$ For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately, then, using this inequality, the bound we get is \begin{align*} &(1+|\sum x_i| + |\sum y_i| + |\sum z_i|)^{p-\delta}\\ &\quad\times \int(1+|\sum(\eta_i^1-\frac13)x_i|+|\sum(\eta_i^2-\frac13)y_i|+|\sum(\eta_i^3-\frac13)z_i|)^p. \end{align*} Use Khinchin's inequality again to show the term on the second line of the above display is $\lesssim_p 1$, finishing the proof.

As for the second part, I think something like this works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C(p)\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to rewrite the left member of (3.4) as $$ \big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big] + R\big|, $$ where $R = \pm\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)\pm\dots$ is the remainder we get by algebraically manipulating this expression to match the given left-hand side of (3.4). Writing (3.4) this way sets us up to use the first part of the proof, and indeed the part not involving $R$ agrees with what Bourgain wrote. To see how to handle $R$, I'll just demonstrate what to do with $\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)$, since the triangle inequality finishes the bound. By Cauchy-Schwarz, \begin{align*} |\E[\phi_1(X)\phi_2(Y)]|\le (\E|\phi_1(X)|^2)^{\frac12}(\E|\phi_2(Y)|^2)^{\frac12}. \end{align*} By the bound (3.1), $|\phi_1(X)|\lesssim (1+|X|)^{p_1}\le(1+|X-\E X|)^{p_1}(1+|\E X|)^{p_1}$, so by Khinchin's inequality, $\E[|\phi_1(X)|^2]\lesssim_{p_1}|x|^{2p_1}(1+|\mathbb E X|)^{2p_1}\le(1+|\E X|)^{2p_1}$. A similar bound applies to $\E|\phi_2(Y)|^2$, of course. To estimate $|\phi_3(\E Z)|$, use (3.1) again to get $(1+|\E Z|)^{p_3}$, and use $p_1+p_2-\delta>0$ to get the final bound $|\phi_3(\E Z)|\lesssim (1+|\sum z_i|)^{p-\delta}$, which is smaller than the right-hand side of (3.4). (Note how the constant doesn't depend on $n$, the number of terms, because we used Khinchin's inequality.)

The last thing to check is $$\big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big]\big|\le \text{RHS (3.4)}.$$ For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately, then, using this inequality, the bound we get is \begin{align*} &(1+|\sum x_i| + |\sum y_i| + |\sum z_i|)^{p-\delta}\\ &\quad\times \int(1+|\sum(\eta_i^1-\frac13)x_i|+|\sum(\eta_i^2-\frac13)y_i|+|\sum(\eta_i^3-\frac13)z_i|)^p. \end{align*} Use Khinchin's inequality again to show the term on the second line of the above display is $\lesssim_p 1$, finishing the proof.

As for the second part, I think something like this works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to rewrite the left member of (3.4) as $$ \big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big] + R\big|, $$ where $R = \pm\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)\pm\dots$ is the remainder we get by algebraically manipulating this expression to match the given left-hand side of (3.4). Writing (3.4) this way sets us up to use the first part of the proof, and indeed the part not involving $R$ agrees with what Bourgain wrote. If you crudely use the triangle inequality on $R$ and apply (3.1) to each of the $6$ terms of $R$, you will end up with, say \begin{align*} &6C(1+|\sum x_i|)^{p_1}(1+|\sum y_i|)^{p_2}(1+|\sum z_i|)^{p_3} \\ &\qquad\le 6C(1+|\sum x_i|+|\sum y_i| + |\sum z_i|)^{p}\\ &\qquad\le 6C'(1+|\sum x_i|+|\sum y_i| + |\sum z_i|)^{p-\delta} \end{align*} for a different $C'$ that may depend on $p,\delta$. This uses that $|x|,|y|,|z|\le 1$, so all expressions $(1+|\sum x_i|+|\sum y_i|+|\sum z_i|)^q$ are comparable for $q \ge 0$ with constants depending on $q$.

The last thing to check is that after integrating the last displayed expression on p. 235 in $t$ (taking expectation), we get $C(1+|\sum x_i|+|\sum y_i| + |\sum z_i|)^{p-\delta}$. Firstly, the terms $$ |\sum (\eta_i^1-\frac13)x_i|,\ |\sum (\eta_i^2-\frac13)y_i|,\ |\sum (\eta_i^3-\frac13)z_i| $$ can each be bounded by $1$ independently of $t$ (again by the assumption $|x|,|y|,|z|\le 1$), and we are left with, say \begin{align*} &C(4+|\sum x_i|+|\sum y_i| + |\sum z_i|)^{p-\delta}\\ &\qquad\times\int(|\sum (\eta_i^1-\frac13)x_i| + |\sum (\eta_i^2-\frac13)y_i|+|\sum(\eta_i^3-\frac13)z_i|)^\delta\,dt. \end{align*} Now we can simply bound the integral by $3^\delta$ to end up with the right member of (3.4).

The only remark I have about this is that it seems like for the applications, it is perfectly alright to allow the constant $C$ to depend on $p,\delta$.

As for the second part, I think something like this works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to rewrite the left member of (3.4) as $$ \big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big] + R\big|, $$ where $R = \pm\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)\pm\dots$ is the remainder we get by algebraically manipulating this expression to match the given left-hand side of (3.4). Writing (3.4) this way sets us up to use the first part of the proof, and indeed the part not involving $R$ agrees with what Bourgain wrote. To see how to handle $R$, I'll just demonstrate what to do with $\E[\phi_1(X)\phi_2(Y)]\phi_3(\E Z)$, since the triangle inequality finishes the bound. By Cauchy-Schwarz, \begin{align*} |\E[\phi_1(X)\phi_2(Y)]|\le (\E|\phi_1(X)^2|)^{\frac12}(\E|\phi_2(Y)^2|)^{\frac12}. \end{align*} By the bound (3.1), $|\phi_1(X)|\lesssim (1+|X|)^{p_1}$, so by Khinchin's inequality, $\E|\phi_1(X)^2|\lesssim_{p_1}|x|^{2p_1}\le1$. A similar bound applies to $\E|\phi_2(Y)|^2$, of course. To estimate $|\phi_3(\E Z)|$, use (3.1) again to get $(1+|\E Z|)^{p_3}$, and use $p_1+p_2-\delta>0$ to get the final bound $|\phi_3(\E Z)|\lesssim (1+|\sum z_i|)^{p-\delta}$, which is smaller than the right-hand side of (3.4). (Note how the constant doesn't depend on $n$, the number of terms, because we used Khinchin's inequality.)

The last thing to check is $$\big|\E\big[(\phi_1(X)-\phi_1(\E X))\cdot(\phi_2(Y)-\phi_2(\E Y))\cdot(\phi_3(Z)-\phi_3(\E Z))\big]\big|\le \text{RHS (3.4)}.$$ For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately, then, using this inequality, the bound we get is \begin{align*} &(1+|\sum x_i| + |\sum y_i| + |\sum z_i|)^{p-\delta}\\ &\quad\times \int(1+|\sum(\eta_i^1-\frac13)x_i|+|\sum(\eta_i^2-\frac13)y_i|+|\sum(\eta_i^3-\frac13)z_i|)^p. \end{align*} Use Khinchin's inequality again to show the term on the second line of the above display is $\lesssim_p 1$, finishing the proof.