# Concentration on the Boolean hypercube via pathwise stochastic analysis

## Abstract

We develop a new technique for proving concentration inequalities which relate the variance and influences of Boolean functions. Using this technique, we

1. 1.

Settle a conjecture of Talagrand (Combinatorica 17(2):275–285, 1997), proving that

\begin{aligned} \int _{\left\{ -1,1\right\} ^{n}}\sqrt{h_{f}\left( x\right) }d\mu \left( x\right) \ge C\cdot \mathrm {Var}\left( f\right) \cdot \left( \log \left( \frac{1}{\sum \mathrm {Inf}_{i}^{2}\left( f\right) }\right) \right) ^{1/2}, \end{aligned}

where $$h_{f}\left( x\right)$$ is the number of edges at x along which f changes its value, $$\mu \left( x\right)$$ is the uniform measure on $$\left\{ -1,1\right\} ^{n}$$, and $$\mathrm {Inf}_{i}\left( f\right)$$ is the influence of the i-th coordinate.

2. 2.

Strengthen several classical inequalities concerning the influences of a Boolean function, showing that near-maximizers must have large vertex boundaries. An inequality due to Talagrand states that for a Boolean function f, $$\mathrm {Var}\left( f\right) \le C\sum _{i=1}^{n}\frac{\mathrm {Inf}_{i}\left( f\right) }{1+\log \left( 1/\mathrm {Inf}_{i}\left( f\right) \right) }$$. We give a lower bound for the size of the vertex boundary of functions saturating this inequality. As a corollary, we show that for sets that satisfy the edge-isoperimetric inequality or the Kahn–Kalai–Linial inequality up to a constant, a constant proportion of the mass is in the inner vertex boundary.

3. 3.

Improve a quantitative relation between influences and noise stability given by Keller and Kindler.

Our proofs rely on techniques based on stochastic calculus, and bypass the use of hypercontractivity common to previous proofs.

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1. We know of no specific Boolean function which satisfies this condition. In fact, we believe that the “large jump” case essentially never happens, i.e. that there exist global constants $$c_{1}$$ and $$c_{2}$$ such that $$\mathbb {P}\left[ f_{\theta }\in \left[ 0,c_{1}\right] \mid \theta <1\right] >c_{2}$$. However, we currently have no proof for such a statement.

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## Acknowledgements

The first author would like to thank Noam Lifshitz for useful discussions and in particular for pointing out the possible application to stability of the isoperimetric inequality. We are also thankful to Ramon Van Handel, Itai Benjamini and Gil Kalai for an enlightening discussion, and to Gregory Rosenthal for his comments. We are grateful to the anonymous referees, whose insightful comments have greatly improved the paper. We thank Mark Sellke for a simplification of the proofs of Lemmas 2.5 and 5.3. R.E. is an incumbent of the Elaine Blond Career Development Chair, and is supported by a European Research Council Starting Grant (ERC StG) and by an Israel Science Foundation Grant No. 718/19. R.G. is supported by the Adams Fellowship Program of the Israel Academy of Sciences and Humanities, the European Research Council and by the Israeli Science Foundation.

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## Appendices

### Appendix A: p-Biased analysis

For $$p=\left( p_{1},\ldots ,p_{n}\right) \in \left[ 0,1\right] ^{n}$$, let $$\mu _{p}$$ be the measure

\begin{aligned} \mu _{p}\left( y\right)&=\prod _{i=1}^{n}\frac{1+y_{i}\left( 2p_{i}-1\right) }{2}=:w_{\left( 2p-1\right) }\left( y\right) , \end{aligned}

which sets the i-th bit to 1 with probability $$p_{i}$$. Let

\begin{aligned} \omega _{i}\left( y\right)&=\frac{1}{2}\left( \frac{1-2p_{i}}{\sqrt{p_{i}\left( 1-p_{i}\right) }}+y_{i}\frac{1}{\sqrt{p_{i}\left( 1-p_{i}\right) }}\right) , \end{aligned}
(95)

and for a set $$S\subseteq \left[ n\right]$$, define $$\omega _{S}\left( y\right) =\prod _{i\in S}\omega _{i}\left( y\right)$$. Then every function f can be written as

\begin{aligned} f\left( y\right)&=\sum _{S\subseteq \left[ n\right] }\hat{f}_{p}\left( S\right) \omega _{S}\left( y\right) \nonumber \\&:=\sum _{S\subseteq \left[ n\right] }\left( \mathbb {E}_{\mu _{p}}\left[ f\cdot \omega _{S}\right] \right) \omega _{S}\left( y\right) \end{aligned}
(96)
\begin{aligned}&=\sum _{S\subseteq \left[ n\right] }\left( \sum _{y\in \left\{ -1,1\right\} ^{n}}f\left( y\right) \omega _{S}\left( y\right) w_{2p-1}\left( y\right) \right) \omega _{S}\left( y\right) . \end{aligned}
(97)

The coefficients $$\hat{f}_{p}\left( S\right) :=\mathbb {E}_{\mu _{p}}\left[ f\cdot \omega _{S}\right]$$ are called the “p-biased” Fourier coefficients of f.

The p-biased influence of the i-th bit is

\begin{aligned} \mathrm {Inf}_{i}^{p}\left( f\right) =4p_{i}\left( 1-p_{i}\right) \mathbb {P}_{y\sim \mu _{p}}\left[ f\left( x\right) \ne f\left( x^{\oplus i}\right) \right] . \end{aligned}

If f is monotone, then

\begin{aligned} \mathrm {Inf}_{i}^{p}\left( f\right) =2\sqrt{p_{i}}\sqrt{1-p_{i}}\hat{f}_{p}\left( \left\{ i\right\} \right) . \end{aligned}
(98)

The p-biased Fourier coefficients are related to the derivatives of f by the following proposition, whose proof (using slightly different notation) can be found in [28, Section 8].

### Proposition A.1

Let $$S=\left\{ i_{1},\ldots ,i_{k}\right\} \subseteq \left[ n\right]$$ be a set of indices, $$x\in \left( -1,1\right) ^{n}$$, and $$p=\frac{1+x}{2}$$. Then

\begin{aligned} \partial _{i_{1}}\ldots \partial _{i_{k}}f\left( x\right) =\left( \prod _{i\in S}\frac{4}{\sqrt{1-x_{i}^{2}}}\right) \hat{f}_{p}\left( S\right) . \end{aligned}

### Proof of Lemma 2.1

We prove this lemma by using the analogous Level-1 inequality for Gaussian sets, which states that among all sets with a fixed Gaussian measure, the center of mass of the Gaussian half-plane is the furthest from the origin. More formally, let $$\gamma _{n}$$ denote the n-dimensional Gaussian measure on $$\mathbb {R}^{n}$$, and let $$A\subseteq \mathbb {R}^{n}$$. Then

\begin{aligned} \left\| \int _{A}z\gamma _{n}\left( dz\right) \right\| _{2}\le \int _{\Phi ^{-1}\left( \gamma _{n}\left( A\right) \right) }^{\infty }t\gamma _{1}\left( dt\right) , \end{aligned}
(99)

where $$\Phi \left( s\right) =\left( 2\pi \right) ^{-1/2}\int _{-\infty }^{s}e^{-z^{2}/2}dz$$ is the one-dimensional Gaussian CDF. For a proof, see [13, Fact 10]. Through a technical calculation, it can be shown [11, Lemma 21] that there exists a constant $$C>0$$ such that for all $$s\in \left( 0,1\right)$$,

\begin{aligned} \int _{\Phi ^{-1}\left( s\right) }^{\infty }t\gamma _{1}\left( dt\right) \le Cs\sqrt{\log \frac{e}{s}}. \end{aligned}
(100)

We now relate the derivative $$\nabla g\left( x\right)$$ to integrals of the form $$\int _{A}z\gamma _{n}\left( dz\right)$$. Let $$p=\frac{1+x}{2}$$, let $$\mu _{p}$$ be the p-biased measure on $$\left\{ -1,1\right\} ^{n}$$, and let $$\alpha \in \mathbb {R}^{n}$$ be defined by $$\alpha _{i}=\Phi ^{-1}\left( 1-p_{i}\right)$$. This means that $$1-p_{i}=\Phi \left( \alpha _{i}\right) =\left( 2\pi \right) ^{-1/2}\int _{-\infty }^{\alpha _{i}}e^{-x^{2}/2}dx$$, and for every function $$f:\left\{ -1,1\right\} ^{n}\rightarrow \mathbb {R}$$,

\begin{aligned} \int _{\mathbb {R}^{n}}f\left( \mathrm {sign}^{n}\left( z+\alpha \right) \right) d\gamma _{n}\left( z\right) =\mathbb {E}_{\mu _{p}}\left[ f\right] =f\left( x\right) . \end{aligned}
(101)

Since g takes the value $$\left\{ 0,1\right\}$$ on $$\left\{ -1,1\right\} ^{n}$$, plugging in $$f=g$$ gives $$\gamma _{n}\left( A\right) =g\left( x\right)$$, where $$A=\left\{ z\in \mathbb {R}^{n}\mid g\left( \mathrm {sign}^{n}\left( z+\alpha \right) \right) =1\right\}$$. For this choice of A, a calculation shows that for all $$i\in \left[ n\right]$$,

\begin{aligned} \int _{A}z_{i}\gamma _{n}\left( dz\right)&=\frac{1}{\sqrt{2\pi }}e^{-\alpha _{i}^{2}/2}\int _{\mathbb {R}^{n}}\partial _{i}g\left( \mathrm {sign}^{n}\left( z+\alpha \right) \right) d\gamma _{n}\left( z\right) \nonumber \\&\overset{(101)}{=}\frac{1}{\sqrt{2\pi }}e^{-\alpha _{i}^{2}/2}\partial _{i}g\left( x\right) \nonumber \\&\ge \frac{1}{\sqrt{2\pi }}e^{-\Phi ^{-1}\left( \frac{1+t}{2}\right) ^{2}/2}\partial _{i}g\left( x\right) . \end{aligned}
(102)

We then have

\begin{aligned} \left\| \nabla g\left( x\right) \right\| _{2}^{2}&\overset{(102)}{\le }2\pi e^{\Phi ^{-1}\left( \frac{1+t}{2}\right) ^{2}}\left\| \int _{A}z\gamma _{n}\left( z\right) \right\| _{2}^{2}\\&\overset{(99)}{\le }2\pi e^{\Phi ^{-1}\left( \frac{1+t}{2}\right) ^{2}}\int _{-\Phi ^{-1}\left( g\left( x\right) \right) }^{\infty }t\gamma _{1}\left( dt\right) \\&\overset{(100)}{\le }Ce^{\Phi ^{-1}\left( \frac{1+t}{2}\right) ^{2}}g\left( x\right) ^{2}\log \frac{e}{g\left( x\right) } \end{aligned}

as needed. $$\square$$

### Proof of Lemma 2.2

Using Proposition A.1 for $$S=\left\{ i,j\right\}$$ and the fact that $$\left| x_{i}\right| =t$$,

\begin{aligned} \left\| \nabla ^{2}g\left( x\right) \right\| _{HS}^{2}&=\sum _{i,j=1}^{n}\left( \partial _{i}\partial _{j}g\left( x\right) \right) ^{2}\nonumber \\&=\sum _{i,j=1}^{n}\left( \frac{16}{1-t^{2}}\hat{g}_{p}\left( \left\{ i,j\right\} \right) \right) ^{2}\nonumber \\&\le 2C\left( t\right) \sum _{S\subseteq \left[ n\right] ,\left| S\right| =2}\hat{g}_{p}\left( S\right) ^{2}. \end{aligned}
(103)

The following lemma, which bounds the sum of squares of p-biased Fourier coefficients, is immediately obtained from the work of Keller and Kindler [22, Lemma 6]. While that work uses hypercontractivity, the lemma can also be proved without hypercontractivity, by following the proof of the non-biased case by Talagrand [33, Theorem 2.4], or Section 3 by Eldan .

### Lemma B.1

Let $$0\le t<1$$, let $$p\in \left( 0,1\right) ^{n}$$ be such that $$p_{i}\in \left\{ \frac{1+t}{2},\frac{1-t}{2}\right\}$$ for all i, and let $$c_{p}=p_{i}\left( 1-p_{i}\right) =\left( 1-t^{2}\right) /4$$. For a function $$g:\left\{ -1,1\right\} ^{n}\rightarrow \left\{ -1,1\right\}$$, let

\begin{aligned} \mathcal {W}\left( f\right) =c_{p}\sum _{i=1}^{n}\mathrm {Inf}_{i}^{p}\left( g\right) ^{2}. \end{aligned}

There exists a function $$C\left( t\right)$$ such that for every g,

\begin{aligned} \sum _{S\subseteq \left[ n\right] ,\left| S\right| =2}\hat{g}_{p}\left( S\right) ^{2}\le C\left( t\right) \mathcal {W}\left( g\right) \cdot \log \left( \frac{2}{\mathcal {W}\left( g\right) }\right) . \end{aligned}
(104)

Combining (103) and (104), we get

\begin{aligned} \left\| \nabla ^{2}g\left( x\right) \right\| _{HS}^{2}\le C\left( t\right) \mathcal {W}\left( g\right) \cdot \log \left( \frac{2}{\mathcal {W}\left( g\right) }\right) . \end{aligned}
(105)

As stated in Eq. (98), for monotone functions the influence of the i-th bit is given by

\begin{aligned} \mathrm {Inf}_{i}^{p}\left( g\right)&=2\sqrt{p_{i}}\sqrt{1-p_{i}}\hat{g}_{p}\left( \left\{ i\right\} \right) \\&=2\sqrt{c_{p}}\hat{g}_{p}\left( \left\{ i\right\} \right) , \end{aligned}

and so,

\begin{aligned} \mathcal {W}\left( f\right) =c_{p}\sum _{i=1}^{n}\mathrm {Inf}_{i}^{p}\left( g\right) ^{2}=4c_{p}^{2}\sum _{i=1}^{n}\hat{g}_{p}\left( \left\{ i\right\} \right) ^{2}. \end{aligned}

On the other hand, using Proposition A.1 with $$S=\left\{ i\right\}$$,

\begin{aligned} \left\| \nabla g\left( x\right) \right\| _{2}^{2}&=\sum _{i=1}^{n}\left( \partial _{i}g\left( x\right) \right) ^{2}\\&=\frac{16}{1-t^{2}}\sum _{i=1}^{n}\hat{g}_{p}\left( \left\{ i\right\} \right) ^{2}\\&=\frac{4}{\left( 1-t^{2}\right) c_{p}^{2}}\mathcal {W}\left( g\right) :=C'\left( t\right) \mathcal {W}\left( g\right) . \end{aligned}

Plugging this into (105), we see that for some $$C\left( t\right)$$ we have

\begin{aligned} \left\| \nabla ^{2}g\left( x\right) \right\| _{HS}^{2}\le C\left( t\right) \left\| \nabla g\left( x\right) \right\| _{2}^{2}\log \left( \frac{C\left( t\right) }{\left\| \nabla g\left( x\right) \right\| _{2}^{2}}\right) . \end{aligned}

$$\square$$

### Proof of Lemma 2.5

Let s be an integer to be chosen later. Let $$c_{2}=\frac{c_{1}}{2\left( 2-\log \left( 1-a\right) ^{2}-\log c_{1}\right) }$$, and suppose for the sake of contradiction that $$\mathbb {P}\left[ \left[ X\right] _{1}\ge c_{1}\left( 1-a\right) ^{2}\right] <c_{2}\left( 1-a\right) ^{2}$$. Then,

\begin{aligned} \mathbb {E}\left[ \left[ X\right] _{1}\right]&\le \mathbb {E}\left[ \left[ X\right] _{1}\mathbf {1}_{\left[ X\right] _{1}\le c_{1}\left( 1-a\right) ^{2}}\right] +\mathbb {E}\left[ \left[ X\right] _{1}\mathbf {1}_{\left[ X\right] _{1}\in \left[ c_{1}\left( 1-a\right) ^{2},s\right] }\right] \\&\quad +\mathbb {E}\left[ \left[ X\right] _{1}\mathbf {1}_{\left[ X\right] _{1}\ge s}\right] \\&\le c_{1}\left( 1-a\right) ^{2}+c_{2}\left( 1-a\right) ^{2}s+\int _{s}^{\infty }\mathbb {P}\left[ \left[ X\right] _{t}\ge t\right] dt. \end{aligned}

To bound the third term, we use the fact that the quadratic variation is sub-exponential, as follows. Denote $$\tau _{0}=0$$, and for every $$k\in \mathbb {N}$$, let $$\tau _{k}:=\inf \left\{ t>0\mid \left[ X\right] _{t}\ge 2k\right\}$$. Recall that the infimum of an empty set is $$\infty$$. Since the total variation is finite almost surely, the random variable $$k^{*}=\sup \left\{ k\in \mathbb {N}\cup \left\{ 0\right\} \mid \tau _{k}<\infty \right\}$$ is finite almost surely. For every measurable stopping time $$\tau \in \left[ 0,1\right]$$, we have $$\mathbb {E}\left[ \left[ X\right] _{1}-\left[ X\right] _{\tau }\right] \le 1$$. Thus

\begin{aligned} \mathbb {P}\left[ k^{*}\ge k+1\mid k^{*}\ge k\right]&=\mathbb {P}\left[ \left[ X\right] _{1}-\left[ X\right] _{\tau _{k}}\ge 2\right] \\&{\mathop {\le }\limits ^{\left( \text {Markov}\right) }}\frac{\mathbb {E}\left[ \left[ X\right] _{1}-\left[ X\right] _{\tau _{k}}\right] }{2}\le \frac{1}{2}. \end{aligned}

By induction we get that

\begin{aligned} \mathbb {P}\left[ \left[ X\right] _{1}\ge 2k\right] =\mathbb {P}\left[ k^{*}\ge k\right] \le 2^{-k}. \end{aligned}

Thus, if $$s\ge 2$$ is even, we have

\begin{aligned} \int _{s}^{\infty }\mathbb {P}\left[ \left[ X\right] _{1}\ge t\right] dt\le 2\cdot \sum _{k=s/2}^{\infty }2^{-k}=4\cdot 2^{-s/2}. \end{aligned}

The expected value of the quadratic variation can then be bounded by

\begin{aligned} \mathbb {E}\left[ \left[ X\right] _{1}\right] \le c_{1}\left( 1-a\right) ^{2}+c_{2}\left( 1-a\right) ^{2}s+4\cdot 2^{-s/2}. \end{aligned}

Choosing s to be the smallest even integer larger than $$2\left( 2-\log _{2}\left( \left( 1-a\right) ^{2}\right) \right. \left. -\log _{2}c_{1}\right)$$, we get

\begin{aligned} \mathbb {E}\left[ \left[ X\right] _{1}\right]&\le c_{1}\left( 1-a\right) ^{2}+4c_{2}\left( 1-a\right) ^{2}\left( 2-\log _{2}\left( \left( 1-a\right) ^{2}\right) -\log _{2}c_{1}\right) \\&\quad +\left( 1-a\right) ^{2}c_{1}\\&=\left( 1-a\right) ^{2}\left( 2c_{1}+4c_{2}\left( 2-\log \left( 1-a\right) ^{2}-\log c_{1}\right) \right) . \end{aligned}

By choice of $$c_{2}=\frac{c_{1}}{2\left( 2-\log \left( 1-a\right) ^{2}-\log c_{1}\right) }$$, we get $$\mathbb {E}\left[ \left[ X\right] _{1}\right] \le 3c_{1}\left( 1-a\right) ^{2}$$, which, since $$c_{1}<1/3$$, contradicts the fact that $$\mathbb {E}\left[ \left[ X\right] _{1}\right] =\mathrm {Var}\left( X_{1}\right) \ge \left( 1-a\right) ^{2}$$. $$\square$$

### Proof of Lemma 3.3

To prove (27), assume first that $$t_{1}>0$$, so that the number of jumps that $$B_{t}$$ makes in the time interval $$\left[ t_{1,}t_{2}\right]$$ is almost surely finite. For any integer $$N>0$$, partition the interval $$\left[ t_{1,}t_{2}\right]$$ into N equal parts, setting $$t_{k}^{N}=t_{1}+\frac{k}{N}\left( t_{2}-t_{1}\right)$$ for $$k=0,\ldots ,N$$. Almost surely, none of the jumps of $$B_{t}$$ occur at any $$t_{k}^{N}$$. Since there are only finitely many jumps, there exists an almost surely finite $$N_{0}$$ so that for all $$N>N_{0}$$, every sub-interval in the partition contains at most one jump point. Since $$g_{t}$$ is left-continuous, we therefore almost surely have

\begin{aligned} \sum _{t\in J_{i}\cap \left[ t_{1,}t_{2}\right] }4t^{2}g_{t}=\lim _{N\rightarrow \infty }\sum _{k=0}^{N-1}4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\mathbf {1}_{J_{i}\cap \left[ t_{k}^{N},t_{k+1}^{N}\right] \ne \emptyset }. \end{aligned}

Since $$g_{t_{k}^{N}}$$ is bounded, the expression

\begin{aligned} \sum _{k=0}^{N-1}4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\mathbf {1}_{J_{i}\cap \left[ t_{k}^{N},t_{k+1}^{N}\right] \ne \emptyset } \end{aligned}

is bounded in absolute value by a constant times the number of jumps of $$B_{t}$$ in the interval $$\left[ t_{1,}t_{2}\right]$$, which is integrable. By the dominated convergence theorem, we then have

\begin{aligned} \mathbb {E}\sum _{t\in J_{i}\cap \left[ t_{1,}t_{2}\right] }4t^{2}g_{t}&=\lim _{N\rightarrow \infty }\mathbb {E}\sum _{k=0}^{N-1}4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\mathbf {1}_{J_{i}\cap \left[ t_{k}^{N},t_{k+1}^{N}\right] \ne \emptyset }. \end{aligned}

Since $$g_{t_{k}^{N}}$$ is measurable with respect to $$\left\{ B_{s}\right\} _{0\le s<t_{k}^{N}}$$, it is independent of whether or not a jump occurred in the interval $$\left[ t_{k}^{N},t_{k+1}^{N}\right]$$, and the expectation breaks up into

\begin{aligned} \mathbb {E}\sum _{t\in J_{i}\cap \left[ t_{1,}t_{2}\right] }4t^{2}g_{t}=\lim _{N\rightarrow \infty }\sum _{k=0}^{N-1}\mathbb {E}\left[ 4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\right] \mathbb {E}\left[ \mathbf {1}_{J_{i}\cap \left[ t_{k}^{N},t_{k+1}^{N}\right] \ne \emptyset }\right] . \end{aligned}
(106)

The set $$J_{i}=\mathrm {Jump}\left( B_{t}^{\left( i\right) }\right)$$ is a Poisson process with rate 1/2t, and so the number of jumps in the interval $$\left[ t_{k}^{N},t_{k+1}^{N}\right]$$ distributes as $$\mathrm {Pois}\left( \lambda \right)$$, where

\begin{aligned} \lambda =\int _{t_{k}^{N}}^{t_{k+1}^{N}}\frac{1}{2t}dt=\frac{1}{2}\log \frac{t_{k+1}^{N}}{t_{k}^{N}}. \end{aligned}

The probability of having at least one jump is then equal to

\begin{aligned} \mathbb {P}\left[ J_{i}\cap \left[ t_{k}^{N},t_{k+1}^{N}\right] \ne \emptyset \right]&=1-e^{-\lambda }=1-\sqrt{\frac{t_{k}^{N}}{t_{k+1}^{N}}}=1-\sqrt{1-\frac{\left( t_{2}-t_{1}\right) /N}{t_{k+1}^{N}}}\\&=\frac{\left( t_{2}-t_{1}\right) /N}{2t_{k+1}^{N}}+O\left( \frac{1}{N^{2}}\right) . \end{aligned}

Plugging this into display (106), we get

\begin{aligned} \mathbb {E}\sum _{t\in J_{i}\cap \left[ t_{1,}t_{2}\right] }4t^{2}g_{t}=\lim _{N\rightarrow \infty }\mathbb {E}\sum _{k=0}^{N-1}4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\left( \frac{\left( t_{2}-t_{1}\right) /N}{2t_{k}^{N}}+O\left( \frac{1}{N^{2}}\right) \right) . \end{aligned}

The factor $$O\left( \frac{1}{N^{2}}\right)$$ is negligible in the limit $$N\rightarrow \infty$$, since the sum contains only N bounded terms. We are left with

\begin{aligned}&\lim _{N\rightarrow \infty }\mathbb {E}\sum _{k=0}^{N-1}4\left( t_{k}^{N}\right) ^{2}g_{t_{k}^{N}}\frac{\left( t_{2}-t_{1}\right) /N}{2t_{k+1}^{N}} =\lim _{N\rightarrow \infty }\mathbb {E}\sum _{k=0}^{N-1}\left[ 2t_{k}^{N}g_{t_{k}^{N}}\right] \frac{t_{2}-t_{1}}{N}\\&\quad \overset{\left( \text {bounded convergence}\right) }{=}\mathbb {E}\lim _{N\rightarrow \infty }\sum _{k=0}^{N-1}\left[ 2t_{k}^{N}g_{t_{k}^{N}}\right] \frac{t_{2}-t_{1}}{N}. \end{aligned}

Since $$g_{t}$$ is continuous almost everywhere, by the definition of the Riemann integral, the limit is equal to $$2\mathbb {E}\int _{t_{1}}^{t_{2}}t\cdot g_{t}dt$$, and we get

\begin{aligned} \mathbb {E}\sum _{t\in J_{i}\cap \left[ t_{1,}t_{2}\right] }4t^{2}g_{t}=2\mathbb {E}\int _{t_{1}}^{t_{2}}t\cdot g_{t}dt \end{aligned}

for all $$t_{1}>0$$. Taking the limit $$t_{1}\rightarrow 0$$ gives the desired result for $$t_{1}=0$$ by continuity of the right hand side in $$t_{1}$$. $$\square$$

### Proof of Lemma 3.7

Let $$t_{1}=\inf \left\{ t\mid g\left( t\right) \ge K\right\}$$, and denote $$L=\max \left\{ x\log \frac{K}{x}\mid x\in \left[ 0,K\right] \right\}$$. Note that L depends only on K. Then, for all $$t\le t_{1}$$, we have

\begin{aligned} g'\left( t\right) \le C\cdot L. \end{aligned}

Integrating, this means that for all $$t\le t_{1}$$

\begin{aligned} g\left( t\right) \le g\left( 0\right) +tCL\le \frac{K}{2}+tCL. \end{aligned}

In particular, $$t_{1}\ge \frac{K}{2CL}$$, otherwise we would have $$g\left( t_{1}\right) <K$$, contradicting the definition of $$t_{1}$$ and continuity of g. Denoting $$t_{0}=\frac{K}{4CL}$$, we must have $$g\left( t\right) <K$$ for all $$t\in \left[ 0,t_{0}\right]$$. This ensures that $$\log \frac{K}{g\left( t\right) }$$ is positive in this interval, which means we can rearrange the differential inequality (36) to give

\begin{aligned} \frac{g'\left( t\right) }{g\left( t\right) \log \frac{K}{g\left( t\right) }}\le C \end{aligned}

for all $$t\in \left[ 0,t_{0}\right]$$. A short calculation reveals that the left hand side is the derivative of $$-\log \log \left( K/g\right)$$. Integrating from 0 to t, we get

\begin{aligned} \log \log \frac{K}{g\left( 0\right) }-\log \log \frac{K}{g\left( t\right) }\le Ct. \end{aligned}

Rearranging gives

\begin{aligned} \log \log \frac{K}{g\left( t\right) }\ge \log \log \frac{K}{g\left( 0\right) }-Ct, \end{aligned}

and exponentiating twice gives the inequality

\begin{aligned} g\left( t\right) \le \left( \frac{1}{K}\right) ^{e^{-Ct}-1}g\left( 0\right) ^{e^{-Ct}}. \end{aligned}

By the Taylor expansion of $$e^{-Ct}$$, we can write $$e^{-Ct}:=1-Ct+R\left( Ct\right)$$ with $$R\left( x\right) =O\left( x^{2}\right)$$ and $$R\left( x\right) >0$$ for $$x>0$$. This gives

\begin{aligned} g\left( t\right)&\le g\left( 0\right) ^{1-Ct+R\left( Ct\right) }\left( \frac{1}{K}\right) ^{-Ct+R\left( Ct\right) }\\&=g\left( 0\right) ^{1-Ct}K^{Ct}\left( \frac{g\left( 0\right) }{K}\right) ^{R\left( Ct\right) }\\&\le g\left( 0\right) ^{1-Ct}K^{Ct}\left( \frac{1}{2}\right) ^{R\left( Ct\right) }\\&\le g\left( 0\right) ^{1-Ct}K^{Ct} \end{aligned}

as needed. $$\square$$

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