$\require{mathtools}
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Adapted from “Criticality of regular formulas” by Benjamin Rossman.
Say you have some function $f$, and a switching lemma falls from the sky:
Let $(J|z)$ be a $p$-random restriction, then $\Pr[\DT(f_{J|z}) \geq t] \leq (p \lambda)^t$.
What are all the things you can say about $f$ as a function of $\lambda$?
Decision tree size
The switching lemma tells us that after a (sufficiently strong) restriction, even though there are $p n$ variables remaining, the value of the function can be determined with only a couple queries. This means that “on average, we’re almost done after querying $n-p n$ variables”, and so we might reasonably hope that the decision tree size for $f$ is at most $2^{n-pn}$ (instead of the worst case $2^n$).
To make this formal, we need to look at the expected decision tree size. Let $p \ce 1/4\lambda$, then
\[\E_{J,z}[\DTsize(f_{J|z})] \leq \sum_{t=0}^\infty 2^t \cdot \Pr_{J,z}[\DT(f_{J|z}) \geq t] = \sum_{t=0}^\infty 2^{-t} = 2.\]
By Markov’s inequality, this means that $90\%$ of the $J$’s have $\E_z[\DTsize] \leq 20$. Also, roughly half of the $J$’s has at least $p n$ variables, so by a union bound, we can find a $J$ that satisfies both properties.
Then, build a decision tree for $f$ by first querying the $n-|J|$ variables outside of $J$, then picking the smallest possible tree at each leaf. The average size of those subtrees is $O(1)$, so the total size of the tree is
\[2^{n-|J|} \cdot O(1) \leq O\p{2^{n - p n}} = 2^{n-\Omega(n/\lambda)}.\]
Correlation with parity
See more details on correlations bounds against $\AC^0$ in Multi-switching lemma.
Random restrictions are just ways to “zoom in” on a random subcube, so for any set $J$ of stars, the correlation of $f$ with parity is equal to the average correlation of $f_{J|z}$ with parity. But unless $f_{J|z}$ is evasive (i.e. has the maximal decision tree depth $|J|$), the correlation is $0$. Indeed, if even one variable $x_i$ is left unqueried at a leaf, then the correlation within that leaf is zero, because the inputs with $x_i=-1$ will cancel out the inputs with $x_i = +1$. And $f_{J|z}$ is rarely evasive.
Concretely, for $p \ce 1/2\lambda$, we have $\Pr_{J,z}[\DT(f_{J|z}) \geq p n] \leq 2^{-p n}$, and a constant fraction of the $J$’s have size $\geq p n$, so by Markov’s inequality, we can find a $J$ that has at least $p n$ stars and such that
\[\Pr_z[\text{$f_{J|z}$ is evasive}] = \Pr_z[\DT(f_{J|z}) = |J|] \leq \Pr_z[\DT(f_{J|z}) \geq p n] \leq O(2^{-p n}).\]
Therefore, the correlation is
\[
|\inner{f,\Par_n}| = \abs{\E_z[\inner{f_{J|z}, \Par_n}]} \leq \Pr_z[\text{$f_{J|z}$ is evasive}] \leq O(2^{-p n}) = 2^{-\Omega(n/\lambda)}
\]
(where the first inequality holds because correlations have absolute value $\leq 1$).
Low degree and concentration
Another nice thing about decision trees is that they have low degree: indeed, you can represent a decision tree by taking the sum of “indicator” functions corresponding to all of its leaves that are labeled $1$. For example, a leaf where the fixed variables are $x_2 \ce -1$, $x_3 \ce +1$ and $x_5 \ce -1$ can be represented as
\[\p{\frac{1-x_2}2}\p{\frac{1+x_3}2}\p{\frac{1-x_5}2}.\]
So a decision tree of depth $t$ can be written as the sum of $2^t$ such “indicators”, each of degree $t$, which means a direct corollary of the switching lemma is
\[
\Pr[\deg(f_{J|z}) \geq t] \leq (p \lambda)^t,
\]
(we just replaced $\DT(\cdot)$ with $\deg(\cdot)$).
Fourier tails
Setting $p \ce 1/2\lambda$, this tells us $\Pr[\deg(f_{J|z}) \geq t] \leq 2^{-t}$: that $f$ is “basically always constant degree” after the restriction. On the other hand, random restrictions only reduce degree by a factor $p$, so intuitively, in order to get degree $O(1)$ after the restriction, $f$ must have had degree basically $O(1/p) = O(\lambda)$ originally.
Formally, we have
\[
\E\b{\W^{\geq pk}[f_{J|z}]} \geq \Omega\p{\W^{\geq k}[f]}.
\]
On the other hand, if the degree of $f_{J|z}$ is less than $p k$, then $\W^{\geq p k}[f_{J|z}] = 0$, and the weight is never more than $1$, so
\[
\E\b{\W^{\geq pk}[f_{J|z}]} \leq \Pr[\deg(f_{J|z}) \geq p k] \leq 2^{-pk} = 2^{-\Omega\p{k/\lambda}}.
\]
Combining the two, we get $\W^{\geq k}[f] \leq 2^{-\Omega(k/\lambda)}$, which means that Fourier weight tapers off exponentially above degree $O(\lambda)$.
Note 1: Actually, the connection between switching lemmas and Fourier tails also goes the other way: Avishay Tal showed that if $\W^{\geq k}[f] \leq 2^{-\Omega(k/\lambda)}$, then $\Pr[\deg(f_{J|z}) \geq t] \leq O(p\lambda)^t$.
Note 2: This also directly implies the correlation bounds with parity we got earlier. Since $\Par_n(x) = x_1 \cdots x_n$, the correlation of any function $f$ with $\Par_n$ is exactly its degree-$n$ Fourier coefficient $\Fou{f}([n])$. So we get
\[
\inner{f,\Par_n}^2 = \Fou{f}([n])^2 = \W^{\geq n}[f] \leq 2^{-\Omega(n/\lambda)}.
\]
Fourier 1-norm
If $g$ is a depth-$t$ decision tree, then the “indicator” functions corresponding to each of its leaves (such as $\p{\frac{1-x_2}2}\p{\frac{1+x_3}2}\p{\frac{1-x_5}2}$) have $2^t$ coefficients, each of them $\pm 2^{-t}$, so each indicator contributes at most $1$ to the Fourier 1-norm $\sum_{S \sse [n]} \abs{\Fou{g}(S)}$. Summing up over the $2^t$ indicators, that means that whenever $\DT(f_{J|z}) \leq t$, we have
\[\sum_{S \sse [n]} \abs{\Fou{f_{J|z}}(S)} \leq 2^t.\]
Setting $p \ce 1/4\lambda$, we can make sure that the 1-norm is small on average:
\[
\E\b{\sum_{S \sse [n]} \abs{\Fou{f_{J|z}}(S)}} \leq \sum_{t=0}^\infty \Pr[\DT(f_{J|z}) = t] \cdot 2^t \leq \sum_{t=0}^\infty 4^{-t} \cdot 2^t = 2.
\]
On the other hand, in Random restrictions vs Fourier, we saw that the Fourier coefficients at degree $k$ get damped by a factor $p^k$ on average. When $k$ is small, this is not a lot of damping, so $f$ cannot have had too much Fourier 1-norm at low degrees. Formally,
\[
\al{
\E\b{\sum_{S \sse [n]} \abs{\Fou{f_{J|z}}(S)}}
&= \sum_{S \sse [n]} \E\b{\abs{\Fou{f_{J|z}}(S)}}\\
&\geq \sum_{S \sse [n]} \abs{\E\b{\Fou{f_{J|z}}(S)}}\tag{triangle inequality}\\
&= \sum_{S \sse [n]} \abs{p^{|S|}\Fou{f}(S)}\\
&\geq p^{k} \sum_{\substack{S \sse [n]\\|S| \leq k}}\abs{\Fou{f}(S)}.
}
\]
Combining the two, we get
\[
\sum_{\substack{S \sse [n]\\|S| \leq k}}\abs{\Fou{f}(S)} \leq 2/p^k = O(\lambda)^k.
\]
Fourier concentration
These bounds on Fourier tails and Fourier 1-norm are the perfect recipe for Fourier concentration!
First, the Fourier tails tell us we can discard all the coefficients of degree above $k$ with error only $2^{-\Omega(k/\lambda)}$. Setting $k \ce O(\lambda \log \frac{1}{\eps})$ is enough to make the error $\le \eps/2$.
Now, the remaining coefficients have small Fourier 1-norm:
\[
\sum_{\substack{S \sse [n]\\|S| \leq k}}\abs{\Fou{f}(S)} \leq O(\lambda)^k = \lambda^{O(\lambda \log \frac{1}{\eps})},
\]
so since lower-degree norm gives sparsity, we can find a small subset $\CS \sse \binom{[n]}{\leq k}$ of them that accounts for all but $\leq \eps/2$ of their Fourier weight:
\[
|\CS| \leq \p{\lambda^{O(\lambda \log \frac{1}{\eps})}}^2/(\eps/2) = \lambda^{O(\lambda \log \frac{1}{\eps})}.
\]
So overall, we proved $\eps$-Fourier concentration on $\lambda^{O(\lambda\log \frac{1}{\eps})}$ coefficients. When $f$ is a $w$-DNF, we can set $\lambda \ce O(w)$ and get concentration on $w^{O(w \log \frac{1}{\eps})}$ coefficients: this is Mansour’s theorem!