Proof. If $A = \{i_1,\ldots ,i_d\}$ and $B = \{j_1,\ldots ,j_d\}$ , then an $(A,B)$ -junta f can be written as a function of $e_{i_sj_t}$ , and in particular as a polynomial in these functions. Since $e_{i_sj_{t_1}} e_{i_sj_{t_2}} = e_{i_{s_1}j_t} e_{i_{s_2}j_t} = 0$ , if $t_1 \neq t_2$ and $s_1 \neq s_2$ , it follows that f can be written as a linear combination of functions $1_T$ for $T \subseteq A \times B$ .

Conversely, if $T = \{(a_1,b_1),\ldots ,(a_d,b_d)\}$ , then $1_T = e_{a_1b_1} \cdots e_{a_db_d}$ .

To see the truth of the second part of the lemma, notice that if $|A|=|B|=d$ and $T \subseteq A \times B$ , then $|T| \leqslant d$ , and conversely, if $|T| \leqslant d$ , then $T \subseteq A \times B$ for some $A,B$ such that $|A|=|B|=d$ .

Proof. Let $U_{A,B}$ consist of all functions f satisfying the stated condition (i.e., $f(\pi ) = f(\tau \pi \sigma )$ whenever $\sigma $ fixes A pointwise and $\tau $ fixes B pointwise).

Let $a \in A$ and $b \in B$ . If $\sigma $ fixes a and $\tau $ fixes b, then $\pi (a) = b$ iff $\tau \pi \sigma (a) = b$ , showing that $e_{ab} \in U_{A,B}$ . It follows that $V_{A,B} \subseteq U_{A,B}$ .

In the other direction, let $f \in U_{A,B}$ . Suppose for definiteness that $A = [a]$ and $B = [b]$ . Let $\pi $ be a permutation such that $\pi (1) = 1, \ldots , \pi (t) = t$ , and $\pi (i)> b$ for $i=t+1,\ldots ,a$ . Applying a permutation fixing B pointwise on the left, we turn $\pi $ into a permutation $\pi '$ such that $\pi '(1),\ldots ,\pi '(a) = 1,\ldots ,t,b+1,\ldots ,b+(a-t)$ . Applying a permutation fixing A pointwise on the right, we turn $\pi '$ into the permutation $1,\ldots ,t,b+1,\ldots ,b+(a-t),\ldots ,n,t+1,\ldots ,a$ . This shows that if $\pi _1,\pi _2$ are two permutations satisfying $e_{ab}(\pi _1) = e_{ab}(\pi _2)$ for all $a \in A,b \in B$ , then we can find permutations $\sigma _1,\sigma _2$ fixing A pointwise and permutations $\tau _1,\tau _2$ fixing B pointwise such that $\tau _1 \pi _1 \sigma _1 = \tau _2 \pi _2 \sigma _2$ , and so $f(\pi _1) = f(\pi _2)$ . This shows that $f \in V_{A,B}$ .

Proof. Let S be a set of size t and let $x=\big ((i_{k},j_{k})\big )_{k\in S}\in L^{S}$ . Let $y\sim L^{\left [m\right ]\setminus S}$ be chosen uniformly and let $\sigma $ be the random permutation that our coupling process outputs given $\left (x,y\right )$ . We have

$$\begin{align*}\|g_{S\to x}\|_{2}^{2}=\mathbb{E}_{y}\left(\mathbb{E}_{\sigma}f\left(\sigma\right)\right)^{2}\leqslant\mathbb{E}_{\sigma}\left[f\left(\sigma\right)^{2}\right] \end{align*}$$

by Cauchy–Schwarz. Next, we consider the values of $\sigma \left (i_{k}\right )$ for $k\in S$ , condition on them and denote $T=\left \{ \left (i_{k},\sigma (i_k)\right )\right \} $ . The conditional distribution of $\sigma $ given T is uniform by the symmetry of elements in $\left [n\right ]\setminus \left \{ i_{k}|\,k\in S\right \}$ , so for any permutation $\pi $ on $\left [n\right ]\setminus \left \{ i_{k}|\,k\in S\right \}$ , we have that $\sigma \pi $ has the same probability as $\sigma $ . Also, the collection $\{\sigma \pi \}$ consists of all permutations satisfying T, so

$$\begin{align*}\mathop{\mathbb{E}}\left[f\left(\sigma\right)^{2}\right]= \mathbb{E}_{T}\left[\mathrm{\|}f_{\to T}\|_{2}^{2}\right]\leqslant\max_{T}\|f_{\to T}\|_{2}^{2} \leqslant C^{2|S|}\varepsilon^2.\\[-42pt] \end{align*}$$

Fact 3.2. Suppose that we are given two probability spaces $\left (X,\mu _{X}\right ),\left (Y,\mu _{Y}\right )$ . Suppose further that for each $x\in X$ , we have a distribution $N\left (x\right )$ on Y, such that if we choose $x\sim \mu _{X}$ and $y\sim N\left (x\right )$ , then the marginal distribution of y is $\mu _{Y}$ . Define an operator $\mathrm {T}_{Y\to X}\colon L^{2}\left (Y\right )\to L^{2}\left (X\right )$ by setting

$$\begin{align*}\mathrm{T}_{Y\to X}f\left(x\right)=\mathbb{E}_{y\sim N\left(x\right)}f\left(y\right). \end{align*}$$

Then $\|\mathrm {T}_{Y\to X}f\|_{q}\leqslant \|f\|_{q}$ for each $q\geqslant 1$ .

Proof. Let $f\colon S_n\to \mathbb {R}$ be $\varepsilon $ -global with constant C. By Proposition 3.1, the function $g=\mathrm {T}_{S_{n}\to L^{m}}f$ is also $\varepsilon $ -constant with constant C, and by Fact 3.2, we have

$$\begin{align*}\left\|\mathrm{T}^{\left(\rho\right)}f\right\|_{q}^{q} = \left\|\mathrm{T}_{L^{m}\to S_{n}}\mathrm{T}_{\rho}g\right\|_{q}^{q}\leqslant \left\|\mathrm{T}_{\rho}g\right\|_{q}^{q}. \end{align*}$$

Now by Theorem 2.5 we may upper-bound the last norm by $\varepsilon ^{q-2}\|g\|_{2}^{2}$ , and using Fact 3.2 again, we may bound $\left \|g\right \|_2^2\leqslant \left \|f\right \|_2^2$ .

Proof. Choose $\rho = 1/(400 C^3 q^2)$ and let P be as in Lemma 3.6. Then

$$\begin{align*}\left\|f\right\|_{q}\leqslant\left\|P\left(\mathrm{T}^{\left(\rho\right)}\right)f\right\|_{q}+\frac{1}{\sqrt{n}}\left\|f\right\|_{2}. \end{align*}$$

As for the first term, we have

$$\begin{align*}\left\|\sum_{i=1}^{l}a_{i}\left(\mathrm{T}^{\left(\rho\right)}\right)^{i}f\right\|_{q} \leqslant \sum_{i=1}^{l}\left|a_{i}\right|\left\|\left(\mathrm{T}^{\left(\rho\right)}\right)^{i}f\right\|_{q} \leqslant \left\|P\right\|\left\|\mathrm{T}^{\left(\rho\right)}f\right\|_{q} \leqslant q^{O\left(d^{3}\right)} \left\|\mathrm{T}^{\left(\rho\right)}f\right\|_{q}. \end{align*}$$

To estimate $\left \|\mathrm {T}^{\left (\rho \right )}f\right \|_{q}$ , note first that by Lemma 3.5, f is $\varepsilon $ -global for constant $4^8$ ; thus, given that C is large enough, we may apply Theorem 3.3 to deduce that $\left \|\mathrm {T}^{\left (\rho \right )}f\right \|_{q}\leqslant \varepsilon ^{\frac {q-2}{q}}\|f\|_{2}^{\frac {2}{q}}$ . As $\|f\|_{2}\leqslant \varepsilon $ , we conclude that

$$\begin{align*}\|f\|_{q}\leqslant q^{O\left(d^{3}\right)}\varepsilon^{\frac{q-2}{q}}\|f\|_{2}^{\frac{2}{q}}+\frac{1}{\sqrt{n}}\|f\|_{2}=q^{O\left(d^{3}\right)}\varepsilon^{\frac{q-2}{q}}\|f\|_{2}^{\frac{2}{q}}.\\[-42pt] \end{align*}$$

Proof. If $T\cup S$ is not consistent, then $1_{T}1_{S}=0$ and so $\left \langle v_{T},v_{S}\right \rangle =0$ . Otherwise,

$$\begin{align*}\left\langle v_{T},v_{S}\right\rangle =\frac{\mathbb{E}\left|1_{T\cup S}\right|}{\|1_{T}\|_{2}\|1_{S}\|_{2}}=\frac{\left(n-\left|T\cup S\right|\right)!}{\sqrt{\left(n-\left|T\right|\right)!\left(n-\left|S\right|\right)!}}\leqslant \frac{(n-d-1)!}{(n-d)!} = O\left(\frac{1}{n}\right).\\[-46pt] \end{align*}$$

Proof. Let $x\sim L^{m},y\sim N_{\rho }\left (x\right )$ and let $\sigma _{x},\sigma _{y}\in S_{n}$ be corresponding permutations chosen according to the coupling. We have

$$\begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{T}\right\rangle =\frac{n!}{\left(n-\left|T\right|\right)!}\left\langle \mathrm{T}^{\left(\rho\right)}1_{T},1_{T}\right\rangle , \end{align*}$$

as $\left \|1_T\right \|_2^2=\frac {\left (n-\left |T\right |\right )!}{n!}$ . We now interpret $\left \langle \mathrm {T}^{\left (\rho \right )}1_{T},1_{T}\right \rangle $ as the probability that both $\sigma _{x}$ and $\sigma _{y}$ satisfy the restrictions given by T. For each sequence S over $[2n]$ of size $\left |T\right |$ , consider the event $A_{S}$ that $x_{S}=y_{S}=T$ , while all the coordinates of the vectors $x_{\left [2n\right ]\setminus S},y_{\left [2n\right ]\setminus S}$ do not contradict T nor belong to T. Then

$$\begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{T}\right\rangle \geqslant\sum_{S\text{ a }|T|\text{-sequence of }\left[2n\right]}\Pr\left[A_{S}\right]. \end{align*}$$

Now the probability that $x_{S}=T$ is $\left (\frac {1}{n}\right )^{2\left |T\right |}$ . Conditioned on $x_{S}=T$ , the probability that $y_{S}=T$ is at least $\rho ^{\left |T\right |}$ . When we condition on $x_{S}=y_{S}=T$ , we obtain that the probability that $x_{\left [n\right ]\setminus S}$ and $y_{\left [n\right ]\setminus S}$ do not involve any coordinate contradicting T or in T is at least $\left (1-\frac {2\left |T\right |}{n}\right )^{2n}=2^{-\Theta \left (|T|\right )}$ . Hence, $\Pr \left [A_{S}\right ]\geqslant \left (\frac {1}{n}\right )^{2\left |T\right |}\Omega \left (\rho \right )^{\left |T\right |}$ . So wrapping everything up, we obtain that

$$\begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{T}\right\rangle \geqslant \frac{(2n)!}{(2n-|T|)!}\cdot\frac{n!}{(n-|T|)!}\frac{1}{n^{2|T|}} \Omega(\rho)^{|T|} =\Omega\left(\rho\right)^{\left|T\right|}.\\[-42pt] \end{align*}$$

Proof. Suppose without loss of generality that $\|1_{T}\|_{2}^{2}\leqslant \|1_{S}\|_{2}^{2}$ , so $\left |T\right |\geqslant \left |S\right |$ . Choose $x\sim L^{m},y\sim N_{\rho }\left (x\right )$ and let $\sigma _{x},\sigma _{y}$ by the corresponding random permutations given by the coupling. We have

$$\begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{S}\right\rangle ={\frac{\Pr\left[1_{T}\left(\sigma_{x}\right)=1,1_{S}\left(\sigma_{y}\right)=1\right]}{\sqrt{\mathbb{E}1_{T}\mathbb{E}1_{S}}}}. \end{align*}$$

As the probability in the numerator is at most $\mathbb {E}\left [1_{T}\right ]$ , we have

$$\begin{align*}\left\langle T^{\left(\rho\right)}v_{T},v_{S}\right\rangle \leqslant\sqrt{\frac{\mathbb{E}\left[1_{T}\right]}{\mathbb{E}\left[1_{S}\right]}}=\sqrt{\frac{\left(n-\left|T\right|\right)!}{\left(n-\left|S\right|\right)!}}, \end{align*}$$

and the proposition follows in the case that $\left |S\right |<\left |T\right |$ .

It remains to prove the proposition provided that $\left |S\right |=\left |T\right |$ . Let $\left (i,j\right )\in S\setminus T$ . Note that

$$\begin{align*}\Pr\left[1_{T}\left(\sigma_{x}\right)=1,1_{S}\left(\sigma_{y}\right)=1\right] \leqslant \frac{1}{n}\Pr\left[1_{T}\left(\sigma_{x}\right)=1 \mid \sigma_{y}\left(i\right)=j\right]. \end{align*}$$

Let us condition further on $\sigma _{x}\left (i\right )$ . Conditioned on $\sigma _{x}\left (i\right )=j$ , we have that $\sigma _{x}$ is a random permutation sending i to j, and so $\Pr \left [1_{T}\left (\sigma _{X}\right )=1\right ]$ is either 0 (if $\left (i,j\right )$ contradicts T) or $\frac {\left (n-1-\left |T\right |\right )!}{\left (n-1\right )!}=O\left (\|1_{T}\|_{2}^{2}\right )$ (if $\left (i,j\right )$ is consistent with T).

Conditioned on $\sigma _{x}\left (i\right )\ne j$ (and on $\sigma _{y}\left (i\right )=j$ ), we again obtain that $\sigma _{x}$ is a random permutation that does not send i to j, in which case,

$$\begin{align*}\Pr\left[1_{T}\left(\sigma_{x}\right)=1\right]=\frac{\left(n-\left|T\right|\right)!}{n!-\left(n-1\right)!}=O\left(\|1_{T}\|_{2}^{2}\right) \end{align*}$$

if $\left (i,j\right )$ contradicts T, and

$$\begin{align*}\Pr\left[1_{T}\left(\sigma_{x}\right)=1\right]=\frac{\left(n-\left|T\right|\right)!-\left(n-\left|T\right|-1\right)!}{n!-\left(n-1\right)!}=O\left(\|1_{T}\|_{2}^{2}\right) \end{align*}$$

if $\left (i,j\right )$ is consistent with T. This completes the proof of the lemma.

Proof. Since $\left \{ v_{T}\right \} _{T\subseteq \left [d\right ]^{2}}$ span the space $V_{\left [d\right ],\left [d\right ]}$ of $\left (\left [d\right ],\left [d\right ]\right )$ -juntas by Lemma 2.3, we may write $f=\sum a_{T}v_{T}$ . Now

$$ \begin{align*} \left\langle \mathrm{T}^{\left(\rho\right)}f,f\right\rangle & =\sum_{T}a_{T}^{2}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{T}\right\rangle +\sum_{T\ne S}a_{T}a_{S}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{S}\right\rangle. \end{align*} $$

By Lemma 5.9, we have

$$ \begin{align*} \left|\sum_{T\ne S}a_{T}a_{S}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{S}\right\rangle \right| \leqslant O\left(\sum_{T\ne S}\frac{\left|a_{T}a_{S}\right|}{\sqrt{n}}\right) \leqslant O\left(\frac{1}{\sqrt{n}}\right)\left(\sum_{T}\left|a_{T}\right|\right)^{2} \leqslant\frac{2^{O\left(d^{2}\right)}}{\sqrt{n}}\left(\sum_{T}\left|a_{T}\right|{}^{2}\right), \end{align*} $$

where the last inequality is by Cauchy–Schwarz. However, by Proposition 5.8, we have

$$\begin{align*}\sum_{T}a_{T}^{2}\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{T}\right\rangle \geqslant\rho^{O\left(d\right)}\left(\sum_{T}a_{T}^{2}\right). \end{align*}$$

Using a similar calculation, one sees that

$$\begin{align*}\|f\|_{2}^{2}=\left(1\pm\frac{2^{O\left(d^{2}\right)}}{n}\right)\sum_{T}a_{T}^{2}, \end{align*}$$

so we get that

$$\begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}f,f\right\rangle \geqslant\left(\rho^{O(d)} - \frac{2^{O\left(d^{2}\right)}}{\sqrt{n}}\right)\sum_{T}a_{T}^{2} \geqslant\left(\rho^{O(d)} - \frac{2^{O\left(d^{2}\right)}}{\sqrt{n}}\right)\left\|f\right\|_2^2 \geqslant \rho^{O(d)}\left\|f\right\|_2^2.\\[-47pt] \end{align*}$$

Proof. By Lemma 5.5, each eigenspace $V_{d,\lambda }$ contains a d-junta. Let $f\in V_{d,\lambda }$ be a nonzero d-junta. Then by Proposition 5.10,

$$\begin{align*}\lambda=\frac{\left\langle \mathrm{T}^{\left(\rho\right)}f,f\right\rangle }{\|f\|_{2}^{2}}\geqslant\rho^{O\left(d\right)}.\\[-45pt] \end{align*}$$

Proof. Let $\lambda $ be an eigenvalue of $\mathrm {T}^{\left (\rho \right )}$ and let f be a corresponding eigenfunction in $V_{\left [d\right ],\left [d\right ]}$ . Write

$$\begin{align*}f=\sum a_{S}v_{S}, \end{align*}$$

where the sum is over all $S=\left \{ \left (i_{1},j_{1}\right ),\ldots ,\left (i_{t},j_{t}\right )\right \} \subseteq \left [d\right ]$ . Then $0=\mathrm {T}^{\left (\rho \right )}f-\lambda f$ , but, however, for each set S, we have

$$ \begin{align*}\left\langle \mathrm{T}^{\left(\rho\right)}f-\lambda f,v_{S}\right\rangle & =a_{S}\left(\left\langle \mathrm{T}^{\left(\rho\right)}v_{S},v_{S}\right\rangle -\lambda\right) \pm\sum_{\left|S\right|\ne\left|T\right|}\left|a_{T}\right|\left(\left|\left\langle \mathrm{T}^{\left(\rho\right)}v_{T},v_{S}\right\rangle \right|+\left|\lambda\right|\left|\left\langle v_{T},v_{S}\right\rangle\right|\right)\\ & =a_{S}\left(\lambda_{\left|S\right|}\left(\rho\right)-\lambda\right)\pm O\left(\frac{\sum_{T\ne S}\left|a_{T}\right|}{\sqrt{n}}\right). \end{align*} $$

Thus, for all S, we have that

$$\begin{align*}\left|a_S\right|\left|\lambda_{\left|S\right|}\left(\rho\right)-\lambda\right|\leqslant O\left(\frac{\sum_{T\ne S}\left|a_{T}\right|}{\sqrt{n}}\right). \end{align*}$$

However, choosing S that maximizes $\left |a_{S}\right |$ , we find that $\left |a_{S}\right |\geqslant \frac {\sum _{T\ne S}\left |a_{T}\right |}{2^{d^{2}}}$ , and plugging that into the previous inequality yields that $\left |\lambda _{\left |S\right |}\left (\rho \right )-\lambda \right |\leqslant \frac {O\left (2^{d^{2}}\right )}{\sqrt {n}}\leqslant n^{-0.4}\rho ^{-d}\leqslant n^{-1/3}\lambda _{\left |S\right |}\left (\rho \right )$ , provided that C is sufficiently large.

Proof. Choose $P(z) = 1-\prod _{i=1}^{d}\left (\lambda _{i}^{-1}z-1\right )^{9d}$ , where $\lambda _i = \lambda _i(\rho )$ . Orthogonally decompose $\mathrm {T}^{\left (\rho \right )}$ to write $f=\sum _{\lambda }f^{=\lambda }$ for nonzero orthogonal functions $f^{=\lambda }\in V_{d}$ satisfying $\mathrm {T}^{\left (\rho \right )}f^{=\lambda }=\lambda f^{=\lambda }$ , and let $g=P\left (\mathrm {T}^{\left (\rho \right )}\right )f-f$ . Then $g=\sum _{\lambda }\left (P\left (\lambda \right )-1\right )f^{=\lambda }$ . Therefore,

$$\begin{align*}\|g\|_{2}^{2}=\sum_{\lambda}\left(P\left(\lambda\right)-1\right)^{2}\|f^{=\lambda}\|_{2}^{2}\leqslant\max_{\lambda}(P\left(\lambda\right)-1)^2\|f\|_{2}^{2}. \end{align*}$$

Suppose the maximum is attained at $\lambda _{\star }$ . By Lemma 5.12, there is $i\leqslant d$ such that $\lambda _{\star }=\lambda _{i}(1\pm n^{-\frac {1}{3}})$ , and so

$$\begin{align*}\left|\left(\lambda_i^{-1} \lambda_{\star} - 1\right)^{9d}\right| \leqslant n^{-3d}. \end{align*}$$

For any $j\neq i$ , we have by Proposition 5.10 that $\lambda _{j}\geqslant \rho ^{O(d)}$ , and so

$$\begin{align*}\left|\left(\lambda_i^{-1} \lambda_{\star} - 1\right)^{9d}\right| \leqslant \rho^{-O(d^2)}. \end{align*}$$

Combining the two inequalities, we get that

$$\begin{align*}(1-P(\lambda_{\star}))^2\leqslant \rho^{-O(d^3)} n^{-6d} \leqslant n^{-2d}, \end{align*}$$

where the last inequality follows from the lower bound on n. To finish up the proof then, we must upper-bound $\left \|P\right \|$ , and this is relatively straightforward:

$$\begin{align*}\left\|P\right\|\leqslant 1 + \left\|\prod_{i=1}^{d}\left(\lambda_{i}^{-1}z-1\right)^{9d}\right\| \leqslant 1 + \prod_{i=1}^{d}\left\|\lambda_{i}^{-1}z-1\right\|^{9d} = 1 + \prod_{i=1}^{d}(1+\lambda^{-1})^{9d} \leqslant 1 + \prod_{i=1}^{d}(1+\rho^{-O(d)})^{9d}, \end{align*}$$

which is at most $\rho ^{-O(d^3)}$ . In the second inequality, we used the fact that $\left \|P_1 P_2\right \|\leqslant \left \|P_1\right \|\left \|P_2\right \|$ .

Proof. Let $\rho =\frac {1}{(10\cdot 4^8\cdot q)^{2}}$ . Decomposing f into the $\sum \limits _{\lambda } f_{=\lambda }$ where $T^{(\rho )} f_{=\lambda } = \lambda f_{=\lambda }$ , we may find g of degree d, such that $f=\mathrm {T}^{\left (\rho \right )}g$ – namely, $g = \sum \limits _{\lambda }\lambda ^{-1} f_{=\lambda }$ . By Parseval and Corollary 5.11, we get that $\|g\|_{2}\leqslant \rho ^{-O(d)}\|f\|_{2}$ . Thus, we have that $\left \|f\right \|_q = \left \|T^{(\rho )} g\right \|_q$ , and to upper-bound this norm, we intend to use Theorem 3.3, and for that, we show that g is global with fairly weak parameters.

Let $T\subseteq L$ be consistent of size at most $2d$ . Then

$$\begin{align*}\|g_{\rightarrow T}\|_{2}^2 = \frac{\mathop{\mathbb{E}}_{x} g(x) 1_T(x)}{\mathop{\mathbb{E}}_x[1_T(x)]} \leqslant \sqrt{ \frac{\mathop{\mathbb{E}}_{x} g(x)^2}{\mathop{\mathbb{E}}_x[1_T(x)]}} \leqslant n^{\frac{\left|T\right|}{2}}\|g\|_{2}^{2} \leqslant n^{\frac{\left|T\right|}{2}}\rho^{-O(d)} \|f\|_{2}^{2}, \end{align*}$$

and so g is $(2d,\varepsilon )$ global for $\varepsilon =n^{d/2}\rho ^{-O(d)}\|f\|_{2}$ . Lemma 3.5 now implies that g is $\varepsilon $ -global with constant $4^8$ . By the choice of $\rho $ , we may now use Theorem 3.3 to deduce that

$$\begin{align*}\left\|T^{(\rho)} g\right\|_q\leqslant \varepsilon^{(q-2)/q} \left\|g\right\|_2^{2/q} \leqslant n^{d/2} \rho^{-O(d)}\left\|f\right\|_2 \leqslant n^{d} q^{O(d)} \left\|f\right\|_2.\\[-42pt] \end{align*}$$

Proof of Lemma 3.6.

Let f be a function of degree d. By Lemma 5.13, there exists a P with $\|P\|\leqslant \rho ^{-O\left (d^{3}\right )}$ and $P\left (0\right )=0$ such that the function $g=P\left (\mathrm {T}^{\left (\rho \right )}\right )f-f$ satisfies $\|g\|_{2}\leqslant n^{-2d}\|f\|_{2.}$ By Lemma 5.14, $\|g\|_{q}\leqslant q^{4d}n^{-d}\|f\|_{2}\leqslant \frac {1}{\sqrt {n}}\|f\|_{2}$ , provided that C is sufficiently large, completing the proof.

Proof. The proof is by induction on t.

Fix $t\geqslant 1$ and $f\in V_{t}$ . Then we may write $f(\pi ) = \sum \limits _{I,J\in [n]_t} {a}({I},{J}) 1_{I\rightarrow J}(\pi )$ , where the coefficients satisfy the symmetry property from Definition 6.1.

Throughout the proof, we will change the coefficients in a sequential process and always maintain the form $f = h + \sum \limits _{\left |I\right | = \left |J\right |=t} {b}({I},{J}) 1_{I\rightarrow J}(\pi )$ for $h\in V_{t-1}$ .

Take $r\in [t]$ , and for each $I = {\left \{ i_1,\ldots ,i_t \right \}}$ , $J = {\left \{ j_1,\ldots ,j_t \right \}}$ , define the coefficients

(5) $$ \begin{align} {b}({I},{J}) = {a}({I},{J}) - \frac{1}{n-t+1}\sum\limits_{i\not\in I\setminus{\left\{ i_r \right\}}} {a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J}). \end{align} $$

In Claim 6.3 below, we prove that after making this change of coefficients, we may write $f = h + \sum \limits _{\left |I\right | = \left |J\right |=t} {b}({I},{J}) 1_{I\rightarrow J}(\pi )$ and that the coefficients ${b}({I},{J})$ satisfy all normalising relations that the ${a}({I},{J})$ do, as well as the normalising relations from the first collection in Definition 6.1 for r. We repeat this process for all $r\in [t]$ .

After this process is done, we have $f = h + \sum \limits _{I,J\in [n]_t} {b}({I},{J}) 1_{I\rightarrow J}(\pi )$ , where the coefficients ${a}({I},{J})$ satisfy the first collection of normalising relations from Definition 6.1. We can now perform the analogous process on the J part, and by symmetry obtain that after this process, the second collection of normalising relations in Definition 6.1 hold. One only has to check that this does not destroy the first collection of normalising relations, which we also prove in Claim 6.3.

Finally, we symmetrize f to ensure that it satisfies the symmetry condition. To do so, we replace $b(I,J)$ with the average of $b(\pi (I),\pi (J))$ over all $\pi \in S_t$ , which does not change the function since $1_{I\to J} = 1_{\pi (I)\to \pi (J)}$ .

Proof. We prove each one of the items separately.

Proof of the first item.

Fix $I = {\left \{ i_1,\ldots ,i_{r-1},i_{r+1}\ldots ,i_t \right \}}$ , $J = {\left \{ j_1,\ldots ,j_t \right \}}$ and calculate:

(6) $$ \begin{align} \sum\limits_{i_r\not\in I} {b}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) &= \sum\limits_{i_r\not\in I} \Bigg({a}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) - \frac{1}{n-t+1}\sum\limits_{i\not\in I}{a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J}) \Bigg)\notag\\ &= \sum\limits_{i_r\not\in I} {a}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) -\frac{1}{n-t+1} \sum\limits_{\substack{i_r\not\in I\\i\not\in I}} {a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J}). \end{align} $$

As in the second double sum, for each $i_r$ the coefficient ${a}({{\left \{ i_1,\ldots ,i_{r-1}, i_r, i_{r+1},\ldots ,i_t \right \}}},{J})$ is counted $n-\left |I\right | = n-t+1$ times, we get that the above expression is equal to $0$ .

Proof of the second item.

Fix $r' \neq r$ and suppose ${a}({\cdot },{\cdot })$ satisfy the first set of normalising relations for $r'$ . Without loss of generality, assume $r'<r$ . Let $I = {\left \{ i_1,\ldots ,i_{r'-1},i_{r'+1},\ldots ,i_t \right \}}$ , $J = {\left \{ j_1,\ldots ,j_t \right \}}$ . Below, we let $i, i_{r'}$ be summation indices and we denote $I' = {\left \{ i_1,\ldots ,i_{r'-1},i_{r'},i_{r'+1},\ldots ,i_{r-1},i,i_r,\ldots ,,i_t \right \}}$ . Calculating as in (6):

(7) $$ \begin{align} \sum\limits_{i_{r'}\not\in I} {b}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) &= \sum\limits_{i_{r'}\not\in I} {a}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) - \frac{1}{n-t+1}\sum\limits_{i\not\in I\setminus{\left\{ i_r \right\}}}{a}({I'},{J})\notag\\ &= \sum\limits_{i_{r'}\not\in I} {a}({{\left\{ i_1,\ldots,i_t \right\}}},{J}) -\frac{1}{n-t+1} \sum\limits_{i_{r'}\not\in I} \sum\limits_{i\not\in I\setminus{\left\{ i_{r} \right\}}}{a}({I'},{J}). \end{align} $$

The first sum is $0$ by the assumption of the second item. For the second sum, we interchange the order of summation to see that it is equal to $\sum \limits _{i\not \in I\setminus {\left \{ i_r \right \}}} \sum \limits _{i_{r'}\not \in I} {a}({I'},{J})$ , and note that for each i, the inner sum is $0$ again by the assumption of the second item.

Proof of the third item.

Fix $r'$ and suppose ${a}({\cdot },{\cdot })$ satisfy the second set of normalising relations for $r'$ . Fix $I = \{i_1,\ldots ,i_t\}$ , $J = \{j_1,\ldots ,j_{r'-1},j_{r'+1},\ldots ,j_t\}$ , $I' = \{i_1,\ldots ,i_{r-1},i,i_{r+1},\ldots ,i_t\}$ , $J' = \{j_1,\ldots ,j_t\}$ and calculate

(8) $$ \begin{align} \sum_{j_r \notin J} b(I,J') &= \sum_{j_r \notin J} \left(a(I,J') - \frac{1}{n-t+1} \sum_{i \notin I \setminus \{i_r\}} a(I',J') \right) \notag \\ &= \sum_{j_r \notin J} a(I,J') - \frac{1}{n-t+1} \sum_{i \notin I \setminus \{i_r\}} \sum_{j_r \notin J} a(I',J'). \end{align} $$

Once again, both sums vanish due to the assumption.

Proof of the fourth item.

For $I = {\left \{ i_1,\ldots ,i_t \right \}}$ , $J = {\left \{ j_1,\ldots ,j_t \right \}}$ , denote

$$\begin{align*}{c}({I},{J}) = \frac{1}{n-t+1}\sum\limits_{i\not\in I\setminus{\left\{ i_r \right\}}}{a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J}), \end{align*}$$

so that ${a}({I},{J}) = {b}({I},{J}) + {c}({I},{J})$ . Plugging this into the representation of f, we see that it is enough to prove that $h(\pi ) = \sum \limits _{I,J} {c}({I},{J}) 1_{I\rightarrow J}(\pi )$ is in $V_{t-1}$ . Writing $I' = I\setminus {\left \{ i_r \right \}}$ , $J' = J\setminus {\left \{ j_r \right \}}$ and expanding, we see that

$$ \begin{align*} h(\pi) &= \frac{1}{n-t+1}\sum\limits_{I,J}1_{I\rightarrow J}(\pi)\sum\limits_{i\not\in I\setminus{\left\{ i_r \right\}}}{a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J})\\ &= \frac{1}{n-t+1}\sum\limits_{I',J'} \sum\limits_{i\not\in I', j_r\not\in J'} {a}({{\left\{ i_1,\ldots,i_{r-1}, i, i_{r+1},\ldots,i_t \right\}}},{J}) \sum\limits_{i_r\not\in I'} 1_{I\rightarrow J}(\pi). \end{align*} $$

Noting that $\sum \limits _{i_r\not \in I'} 1_{I\rightarrow J}(\pi ) = 1_{I'\rightarrow J'}(\pi )$ is in the spanning set (4) for $t-1$ , the proof is concluded.

Proof. By symmetry, it suffices to prove the statement for R that are prefixes of I. We prove the claim by induction on t. The case $t=0$ is trivial, so assume the claim holds for $t-1$ , where $t\geqslant 1$ , and prove for t. The left-hand side is equal to

$$\begin{align*}\sum\limits_{\substack{i_1,\ldots,i_t\not\in I\\ \text{distinct}}}{{a}({R\circ (i_1,\ldots,i_t)},{J})}. \end{align*}$$

For fixed $i_1,\ldots ,i_{t-1}\not \in I$ , by the normalising relations, we have that

$$\begin{align*}\sum\limits_{i_t\not\in I\cup{\left\{ i_1,\ldots,i_{t-1} \right\}}}{{a}({R\circ (i_1,\ldots,i_{t-1})\circ (i_t)},{J})} = -\sum\limits_{i_t\in I\setminus R}{{a}({R\circ (i_1,i_2,\ldots,i_{t-1})\circ (i_t)},{J})}. \end{align*}$$

Hence,

$$\begin{align*}\sum\limits_{\substack{i_1,\ldots,i_t\not\in I\\ \text{distinct}}}{ {a}({R\circ (i_1,\ldots,i_t)},{J})} =-\sum\limits_{i_t\in I\setminus R}{ \sum\limits_{\substack{~~~i_1,\ldots,i_{t-1}\not\in I\cup {\left\{ i_t \right\}}\\ \text{distinct}}}{{a}({R\circ (i_1,i_2,\ldots,i_{t-1})\circ i_t},{J})}}. \end{align*}$$

For fixed $i_t\in I\setminus R$ , using the induction hypothesis, the inner sum is equal to

$$\begin{align*}(-1)^{t-1} \sum\limits_{T\in (I\setminus (R\cup{\left\{ i_t \right\}}))_{t-1}}{a}({R\circ T\circ (i_t)},{J}). \end{align*}$$

Plugging that in,

$$ \begin{align*} \sum\limits_{\substack{i_1,\ldots,i_t\not\in I\\ \text{distinct}}}{ {a}({R\circ (i_1,\ldots,i_t)},{J})} &=(-1)^{t-1}~\sum\limits_{i_t\in I\setminus R}{-\sum\limits_{T\in (I\setminus (R\cup{\left\{ i_t \right\}}))_{t-1}}{a}({R\circ T\circ (i_t)},{J})}\\ &=(-1)^t \sum\limits_{T'\in (I\setminus R)_t}{a}({R\circ T'},{J}).\\[-47pt] \end{align*} $$

Proof. Deferred to Section 6.4.1.

Proof. Take T in this sum for which $I_T[g]\neq 0$ and denote $t=\left |T\right |$ . Then $T = {\left \{ (i_1,j_1),\ldots ,(i_t,j_t) \right \}} = S(I',J')$ for $I' = {\left \{ i_1,\ldots ,i_t \right \}}$ , $J'={\left \{ j_1,\ldots ,j_t \right \}}$ that are consistent. For $Q\subseteq [n]\times [n]$ of size d such that $T\subseteq Q$ , let $S_{Q,T} = \left \{ \left. (I,J) \;\right \vert T\subseteq S(I,J) = Q \right \}$ and note that by the symmetry normalising relation, ${a}({I},{J})$ is constant on $(I,J)\in S_{Q,T}$ . We thus get

$$\begin{align*}I_T[g] =\sum\limits_{Q}{\left(\sum\limits_{(I,J)\in S_{Q,T}}\sqrt{p(1-p)}^d{a}({I},{J})\right)^2} \leqslant d! p^d\sum\limits_{Q}{\sum\limits_{(I,J)\in S_{Q,T}}{a}({I},{J})^2}, \end{align*}$$

where we used the fact that the size of $S_{Q,T}$ is $d!$ . Rewriting the sum by first choosing the locations of T in $(I,J)$ , we get that the last sum is at most

$$\begin{align*}d_t \sum\limits_{\substack{I\in ([n]\setminus I')_{d-t}\\J\in ([n]\setminus J')_{d-t}}} {a}({I'\circ I},{J'\circ J})^2. \end{align*}$$

Combining all, we get that $I_T[g]\leqslant \sum \limits _{\substack {I\in ([n]\setminus I')_{d-t}\\J\in ([n]\setminus J')_{d-t}}} d!^2\frac {1}{n^d} {a}({I'\circ I},{J'\circ J})^2 = I_{I',J'}[g]$ .

Proof. Since the proof is straightforward, we only outline its steps. Writing $f = f_{0}+\dots +f_{d}$ for $f_{k}\in V_{k}$ given by normalising relations, one bounds $\left \|f\right \|_4^4\leqslant (d+1)^3 \sum \limits _{k=0}^{d} \left \|f_{k}\right \|_4^4$ and uses Theorem 6.8 on each $f_{k}$ . Finally, $I_{I',J'}[f_{k}]\leqslant I_{I',J'}[f]\leqslant \varepsilon $ .

Proof. Deferred to Section 8.