Jordan Bell
July 16, 2014

## 1 Binary expansions

Define $S:\{0,1\}^{\mathbb{N}}\to[0,1]$ by

 $S(\sigma)=\sum_{k=1}^{\infty}\frac{\sigma_{k}}{2^{k}},\qquad\sigma\in\{0,1\}^{% \mathbb{N}}.$

For example, for $\sigma_{1}=0$ and $\sigma_{2}=1,\sigma_{3}=1,\ldots$,

 $S(\sigma)=\frac{0}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=\frac{1}{2};$

for $\sigma_{1}=1$ and $\sigma_{2}=0$, $\sigma_{3}=0,\ldots$,

 $S(\sigma)=\frac{1}{2}+\frac{0}{4}+\frac{0}{8}+\cdots=\frac{1}{2}.$

Let $\sigma\in\{0,1\}^{\mathbb{N}}$. If there is some $n\in\mathbb{N}$ such that $\sigma_{n}=0$ and $\sigma_{k}=1$ for all $k\geq n+1$, then defining

 $\tau_{k}=\begin{cases}\sigma_{k}&k\leq n-1\\ 1&k=n\\ 0&k\geq n,\end{cases}$

we have

 $S(\sigma)=\sum_{k=1}^{n-1}\frac{\sigma_{k}}{2^{k}}+\sum_{k=n+1}^{\infty}\frac{% 1}{2^{k}}=\sum_{k=1}^{n-1}\frac{\sigma_{k}}{2^{k}}+\frac{1}{2^{n}}=S(\tau).$

One proves that if either (i) there is some $n\in\mathbb{N}$ such that $\sigma_{n}=0$ and $\sigma_{k}=1$ for all $k\geq n+1$ or (ii) there is some $n\in\mathbb{N}$ such that $\sigma_{n}=1$ and $\sigma_{k}=0$ for all $k\geq n+1$, then $S^{-1}(S(\sigma))$ contains exactly two elements, and that otherwise $S^{-1}(S(\sigma))$ contains exactly one element.

In words, except for the sequence whose terms are only 0 or the sequence whose terms are only 1, $S^{-1}(S(\sigma))$ contains exactly two elements when $\sigma$ is eventually $0$ or eventually $1$, and $S^{-1}(S(\sigma))$ contains exactly one element otherwise.

We define $\epsilon:[0,1]\to\{0,1\}^{\mathbb{N}}$ by taking $\epsilon(t)$ to be the unique element of $S^{-1}(t)$ if $S^{-1}(t)$ contains exactly one element, and to be the element of $S^{-1}(t)$ that is eventually $0$ if $S^{-1}(t)$ contains exactly two elements. For $k\in\mathbb{N}$ we define $\epsilon_{k}:[0,1]\to\{0,1\}$ by

 $\epsilon_{k}(t)=\epsilon(t)_{k},\qquad t\in[0,1].$

Then, for all $t\in[0,1]$,

 $t=S(\epsilon(t))=\sum_{k=1}^{\infty}\frac{\epsilon_{k}(t)}{2^{k}},$ (1)

which we call the binary expansion of $t$.

For $k\in\mathbb{N}$, the $k$th Rademacher function $r_{k}:[0,1]\to\{-1,+1\}$ is defined by

 $r_{k}(t)=1-2\epsilon_{k}(t),\qquad t\in[0,1].$

We can rewrite the binary expansion of $t\in[0,1]$ in (1) as

 $\sum_{k=1}^{\infty}\frac{r_{k}(t)}{2^{k}}=\sum_{k=1}^{\infty}\left(\frac{1}{2^% {k}}-2\cdot\frac{\epsilon_{k}(t)}{2^{k}}\right)=1-2\sum_{k=1}^{\infty}\frac{% \epsilon_{k}(t)}{2^{k}}=1-2t.$ (2)

Define $r:\mathbb{R}\to\{-1,+1\}$ by

 $r(x)=(-1)^{[x]},$

where $[x]$ denotes the greatest integer $\leq x$. Thus, for $0\leq x<1$ we have $r(x)=1$, for $1\leq x<2$ we have $r(x)=-1$, and $r$ has period $2$.

###### Lemma 1.

For any $n\in\mathbb{N}$,

 $r_{n}(t)=(-1)^{[2^{n}t]}=r(2^{n}t),\qquad t\in[0,1]$

In the following theorem we use the Rademacher functions to prove an identity for trigonometric functions.11 1 Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, p. 4, §3.

###### Theorem 2.

For any nonzero real $x$,

 $\prod_{k=1}^{\infty}\cos\frac{x}{2^{k}}=\frac{\sin x}{x}.$
###### Proof.

Let $n\in\mathbb{N}$ and let $c_{1},\ldots,c_{n}\in\mathbb{R}$. The function

 $\sum_{k=1}^{n}c_{k}r_{k}$

is constant on each of the intervals

 $\left[\frac{s}{2^{n}},\frac{s+1}{2^{n}}\right),\qquad 0\leq s\leq 2^{n}-1.$ (3)

There is a bijection between $\Delta_{n}=\{-1,+1\}^{n}$ and the collection of intervals (3). Without explicitly describing this bijection, we have

 $\displaystyle\int_{0}^{1}\exp\left(i\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt$ $\displaystyle=$ $\displaystyle\sum_{s=0}^{2^{n}-1}\int_{s\cdot 2^{-n}}^{(s+1)\cdot 2^{-n}}\exp% \left(i\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt$ $\displaystyle=$ $\displaystyle\sum_{\delta\in\Delta_{n}}\frac{1}{2^{n}}\exp\left(i\sum_{k=1}^{n% }\delta_{k}c_{k}\right)$ $\displaystyle=$ $\displaystyle\sum_{\delta\in\Delta_{n}}\prod_{k=1}^{n}\frac{e^{i\delta_{k}c_{k% }}}{2}$ $\displaystyle=$ $\displaystyle\prod_{k=1}^{n}\frac{e^{ic_{k}}+e^{-ic_{k}}}{2},$

giving

 $\int_{0}^{1}\exp\left(i\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt=\prod_{k=1}^{n}% \cos c_{k}.$ (4)

We have

 $\int_{0}^{1}e^{ix(1-2t)}dt=e^{ix}\frac{e^{-2ixt}}{-2ix}\bigg{|}_{0}^{1}=e^{ix}% \left(\frac{e^{-2ix}}{-2ix}+\frac{1}{2ix}\right)=\frac{\sin x}{x}.$ (5)

Using (2) we check that the sequence of functions $\sum_{k=1}^{n}\frac{r_{k}(t)}{2^{k}}$ converges uniformly on $[0,1]$ to $1-2t$, and hence using (5) we get

 $\int_{0}^{1}\exp\left(ix\sum_{k=1}^{n}\frac{r_{k}(t)}{2^{k}}\right)dt\to\int_{% 0}^{1}e^{ix(1-2t})dt=\frac{\sin x}{x}$

as $n\to\infty$. Combining this with (4), which we apply with $c_{k}=\frac{x}{2^{k}}$, we get

 $\prod_{k=1}^{n}\cos\frac{x}{2^{k}}\to\frac{\sin x}{x}$

as $n\to\infty$, proving the claim. ∎

We now give an explicit formula for the measure of those $t$ for which exactly $l$ of $r_{1}(t),\ldots,r_{n}(t)$ are equal to $1$.22 2 Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, pp. 8–9. We denote by $\mu$ Lebesgue measure on $\mathbb{R}$. We can interpret the following formula as stating the probability that out of $n$ tosses of a coin, exactly $l$ of the outcomes are heads.

###### Theorem 3.

For $n\in\mathbb{N}$ and $0\leq l\leq n$,

 $\mu\{t\in[0,1]:r_{1}(t)+\cdots+r_{n}(t)=2l-n\}=\frac{1}{2^{n}}\binom{n}{l}.$
###### Proof.

Define $\phi:[0,1]\to\mathbb{R}$ by

 $\phi(t)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{ix\left(-(2l-n)+\sum_{k=1}^{n}r_{k}(t)% \right)}dx$

But for $m\in\mathbb{Z}$,

 $\frac{1}{2\pi}\int_{0}^{2\pi}e^{imx}dx=\delta_{m,0}=\begin{cases}1&m=0\\ 0&m\neq 0,\end{cases}$ (6)

hence

 $\phi(t)=\begin{cases}1&\sum_{k=1}^{n}r_{k}(t)=2l-n\\ 0&\sum_{k=1}^{n}r_{k}(t)\neq 2l-n.\end{cases}$

Therefore

 $\displaystyle\mu\bigg{\{}t\in[0,1]:\sum_{k=1}^{n}r_{k}(t)=2l-n\bigg{\}}$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\phi(t)dt$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\frac{1}{2\pi}\int_{0}^{2\pi}e^{ix\left(-(2l-n)+\sum_% {k=1}^{n}r_{k}(t)\right)}dxdt$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ix(2l-n)}\int_{0}^{1}e^{ix\sum_{% k=1}^{n}r_{k}(t)}dtdx$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ix(2l-n)}\cos^{n}xdx;$

the last equality uses (4) with $c_{1}=x,\ldots,c_{n}=x$. Furthermore, writing

 $\cos^{n}x=2^{-n}(e^{ix}+e^{-ix})=2^{-n}\sum_{k=0}^{n}\binom{n}{k}e^{ix(2k-n)},$

we calculate using (6) that

 $\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ix(2l-n)}\cos^{n}xdx$ $\displaystyle=$ $\displaystyle 2^{-n}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2\pi}\int_{0}^{1}e^{-ix% (2l-n)}e^{ix(2k-n)}dx$ $\displaystyle=$ $\displaystyle 2^{-n}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2\pi}\int_{0}^{1}e^{ix(% 2k-2l)}dx$ $\displaystyle=$ $\displaystyle 2^{-n}\sum_{k=0}^{n}\binom{n}{k}\delta_{2k-2l,0}$ $\displaystyle=$ $\displaystyle 2^{-n}\sum_{k=0}^{n}\binom{n}{k}\delta_{k,l}$ $\displaystyle=$ $\displaystyle 2^{-n}\binom{n}{l},$

proving the claim. ∎

We now prove that the expected value of a product of distinct Rademacher functions is equal to the product of their expected values.33 3 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 185, Solution 13.2.

###### Theorem 4.

If $k_{1},\ldots,k_{n}$ are positive integers and $k_{1}<\cdots, then

 $\int_{0}^{1}r_{k_{1}}(t)\cdots r_{k_{n}}(t)dt=0.$
###### Proof.

Write $J=\int_{0}^{1}r_{k_{1}}(t)\cdots r_{k_{n}}(t)dt$ and define

 $\phi(x)=\prod_{s=2}^{n}r(2^{k_{s}-k_{1}}x),\qquad x\in\mathbb{R},$

which satisfies

 $\phi(x+1)=\prod_{s=2}^{n}r(2^{k_{s}-k_{1}}x+2^{k_{s}-k_{1}})=\prod_{s=2}^{n}r(% 2^{k_{s}-k_{1}}x)=\phi(x).$

Hence, as $\phi$ has period $1$ and $r$ has period $2$,

 $\displaystyle J$ $\displaystyle=$ $\displaystyle\int_{0}^{1}r_{k_{1}}(t)\phi(2^{k_{1}}t)dt$ $\displaystyle=$ $\displaystyle\int_{0}^{1}r(2^{k_{1}}t)\phi(2^{k_{1}}t)dt$ $\displaystyle=$ $\displaystyle\frac{1}{2^{k_{1}}}\int_{0}^{2^{k_{1}}}r(x)\phi(x)dx$ $\displaystyle=$ $\displaystyle\frac{1}{2^{k_{1}}}\sum_{j=0}^{2^{k_{1}-1}-1}\int_{2j}^{2j+2}r(x)% \phi(x)dx$ $\displaystyle=$ $\displaystyle\frac{1}{2^{k_{1}}}\sum_{j=0}^{2^{k_{1}-1}-1}\int_{0}^{2}r(x)\phi% (x)dx$ $\displaystyle=$ $\displaystyle\frac{1}{2}\int_{0}^{2}r(x)\phi(x)dx.$

But, as $\phi$ has period $1$,

 $\int_{0}^{2}r(x)\phi(x)dx=\int_{0}^{1}\phi(x)dx-\int_{1}^{2}\phi(x)dx=\int_{0}% ^{1}\phi(x)dx-\int_{0}^{1}\phi(x)dx=0,$

hence $J=0$, proving the claim. ∎

For each $n\in\mathbb{N}$, if $f$ is a function defined on the integers we define

 $I_{n}(f)=\int_{0}^{1}f\left(\sum_{k=1}^{n}r_{k}(t)\right)dt.$
###### Lemma 5.

For any $n\in\mathbb{N}$,

 $I_{n}(x^{2})=n,\qquad I_{n}(x^{4})=3n^{2}-2n.$
###### Proof.

Using Theorem 4 we get

 $\displaystyle I_{n}(x^{2})$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\left(\sum_{k=1}^{n}r_{k}(t)\right)^{2}dt$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\sum_{k=1}^{n}r_{k}(t)^{2}+\sum_{j\neq k}r_{j}(t)r_{k% }(t)dt$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\sum_{k=1}^{n}r_{k}(t)^{2}dt$ $\displaystyle=$ $\displaystyle n.$

Using Theorem 4 we get, since $r_{j}(t)^{4}=r_{j}(t)^{2}=1$ and $r_{j}(t)^{3}=r_{j}(t)$,

 $\displaystyle I_{n}(x^{4})$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\left(\sum_{k=1}^{n}r_{k}(t)\right)^{4}dt$ $\displaystyle=$ $\displaystyle\int_{0}^{1}\sum_{k=1}^{n}r_{k}(t)^{4}+\binom{4}{3}\sum_{j=1}^{n}% \sum_{k\neq j}r_{j}(t)^{3}r_{k}(t)+\binom{4}{2}\sum_{j=1}^{n}\sum_{k\neq j}r_{% j}(t)^{2}r_{k}(t)^{2}$ $\displaystyle+\binom{4}{2}\sum_{j=1}^{n}\sum_{\textrm{j,k,l all distinct}}r_% {j}(t)^{2}r_{k}(t)r_{l}(t)$ $\displaystyle+\sum_{\textrm{j,k,l,m all distinct}}r_{j}(t)r_{k}(t)r_{l}(t)r_% {m}(t)dt$ $\displaystyle=$ $\displaystyle n+\binom{4}{2}n(n-1).$

Our proof of the next identity follows Hata.44 4 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 188, Solution 13.6.

###### Lemma 6.

For any $n\in\mathbb{N}$,

 $I_{n}(|x|)=\frac{2}{\pi}\int_{0}^{\infty}\frac{1-\cos^{n}x}{x^{2}}dx.$
###### Proof.

For $n\in\mathbb{N}$ and $c_{1},\ldots,c_{n}\in\mathbb{R}$,

 $\int_{0}^{1}\exp\left(i\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt=\int_{0}^{1}\cos% \left(\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt+i\int_{0}^{1}\sin\left(\sum_{k=1}^{% n}c_{k}r_{k}(t)\right)dt,$

and since (4) tells us that the left-hand side of the above is real, it follows that we can write (4) as

 $\int_{0}^{1}\cos\left(\sum_{k=1}^{n}c_{k}r_{k}(t)\right)dt=\prod_{k=1}^{n}\cos c% _{k}.$ (7)

Suppose that $\xi$ is a positive real number. Using $t=x\xi$ and doing integration by parts,

 $\displaystyle\int_{0}^{\infty}\frac{1-\cos x\xi}{x^{2}}dx$ $\displaystyle=$ $\displaystyle\xi\int_{0}^{\infty}\frac{1-\cos t}{t^{2}}dt$ $\displaystyle=$ $\displaystyle\xi\frac{1-\cos t}{-t}\bigg{|}_{0}^{\infty}+\xi\int_{0}^{\infty}% \frac{\sin t}{t}dt$ $\displaystyle=$ $\displaystyle\xi\int_{0}^{\infty}\frac{\sin t}{t}dt$ $\displaystyle=$ $\displaystyle\xi\frac{\pi}{2}.$

It is thus apparent that for any real $\xi$,

 $\int_{0}^{\infty}\frac{1-\cos x\xi}{x^{2}}dx=|\xi|\frac{\pi}{2}.$

For any $n\in\mathbb{N}$, applying the above with $\xi=\sum_{k=1}^{n}r_{k}(t)$ we get

 $\displaystyle I_{n}(|x|)$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\int_{0}^{1}\left|\sum_{k=1}^{n}r_{k}(t)\right|\frac% {\pi}{2}dt$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\int_{0}^{1}\int_{0}^{\infty}\frac{1-\cos\left(x\sum% _{k=1}^{n}r_{k}(t)\right)}{x^{2}}dxdt$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{x^{2}}\int_{0}^{1}1-\cos% \left(x\sum_{k=1}^{n}r_{k}(t)\right)dtdx$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{x^{2}}\left(1-I_{n}(\cos x% \cdot)\right)dx.$

Applying (7) with $c_{k}=x$ for each $k$, this is equal to

 $\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{x^{2}}\left(1-\prod_{k=1}^{n}\cos x% \right)dx=\frac{2}{\pi}\int_{0}^{\infty}\frac{1-\cos^{n}x}{x^{2}}dx,$

completing the proof. ∎

We use the above formula for $I_{n}(|x|)$ to obtain an asymptotic formula for $I_{n}(|x|)$.55 5 Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, p. 12, Masayoshi Hata, Problems and Solutions in Real Analysis, p. 188, Solution 13.6.

###### Theorem 7.
 $I_{n}(|x|)\sim\sqrt{\frac{2}{\pi}}\sqrt{n}.$
###### Proof.

By Lemma 6,

 $I_{n}(|x|)=\frac{2}{\pi}\int_{0}^{\infty}\frac{1-\cos^{n}x}{x^{2}}dx.$

For $0\leq\epsilon<1$, define $\phi_{\epsilon}:\left[0,\frac{\pi}{2}\right)\to\mathbb{R}$ by

 $\phi_{\epsilon}(x)=\frac{x^{2}}{2(1-\epsilon)}+\log\cos x.$

We also define

 $\alpha_{\epsilon}=\arccos\sqrt{1-\epsilon},\qquad\beta_{\epsilon}=\int_{\alpha% _{\epsilon}}^{\infty}\frac{1-\cos^{n}x}{x^{2}}dx,$

and for $\sigma>0$,

 $K_{\epsilon,\sigma}=\int_{0}^{\alpha_{\epsilon}}\frac{1-\exp\left(-\frac{nx^{2% }}{\sigma}\right)}{x^{2}}dx.$

Let $0<\epsilon<1$. Until the end of the proof, at which point we take $\epsilon\to 0$, we shall keep $\epsilon$ fixed. For $0 we have, using $\arccos\sqrt{1-\epsilon}\leq\sqrt{\epsilon}$,

 $\phi_{0}(x)=\frac{x^{2}}{2}+\log\cos x<\frac{\epsilon}{2}+\log\sqrt{1-\epsilon% }=\frac{\epsilon}{2}+\frac{1}{2}\log(1-\epsilon)<0,$

hence

 $\cos x<\exp\left(-\frac{x^{2}}{2}\right).$

On the other hand,

 $\phi_{\epsilon}^{\prime}(x)=\frac{x}{1-\epsilon}-\tan x,\qquad\phi_{\epsilon}^% {\prime\prime}(x)=\frac{1}{1-\epsilon}-\sec^{2}x,$

so $\phi_{\epsilon}(0)=\phi_{\epsilon}^{\prime}(0)=0$ and $\phi_{\epsilon}^{\prime\prime}(t)>0$ for all $0\leq t<\alpha_{\epsilon}$, giving

 $\phi_{\epsilon}(x)>0,$

and hence

 $\exp\left(-\frac{x^{2}}{2(1-\epsilon)}\right)<\cos x.$

Collecting what we have established so far, for $0 we have

 $\exp\left(-\frac{x^{2}}{2(1-\epsilon)}\right)<\cos x<\exp\left(-\frac{x^{2}}{2% }\right).$

This shows that

 $K_{\epsilon,2(1-\epsilon)}=\int_{0}^{\alpha_{\epsilon}}\frac{1-\exp\left(-% \frac{nx^{2}}{2(1-\epsilon)}\right)}{x^{2}}dx\geq\int_{0}^{\alpha_{\epsilon}}% \frac{1-\cos^{n}x}{x^{2}}dx,$

and therefore

 $K_{\epsilon,2(1-\epsilon)}+\beta_{\epsilon}\geq\frac{\pi}{2}I_{n}(|x|).$

On the other hand,

 $K_{\epsilon,2}=\int_{0}^{\alpha_{\epsilon}}\frac{1-\exp\left(-\frac{nx^{2}}{2}% \right)}{x^{2}}dx\leq\int_{0}^{\alpha_{\epsilon}}\frac{1-\cos^{n}x}{x^{2}}dx,$

so

 $K_{\epsilon,2}+\beta_{\epsilon}\leq\frac{\pi}{2}I_{n}(|x|).$

Now summarizing what we have obtained, we have

 $K_{\epsilon,2}+\beta_{\epsilon}\leq\frac{\pi}{2}I_{n}(|x|)\leq K_{\epsilon,2(1% -\epsilon)}+\beta_{\epsilon}.$ (8)

For $\sigma>0$, doing the change of variable $t=\sqrt{\frac{n}{\sigma}}x$,

 $K_{\epsilon,\sigma}=\int_{0}^{\alpha_{\epsilon}}\frac{1-\exp\left(-\frac{nx^{2% }}{\sigma}\right)}{x^{2}}dx=\sqrt{\frac{n}{\sigma}}\int_{0}^{\sqrt{\frac{n}{% \sigma}}\alpha_{\epsilon}}\frac{1-e^{-t^{2}}}{t^{2}}dt.$

As $n\to\infty$, the right-hand side of this is asymptotic to

 $\sqrt{\frac{n}{\sigma}}\int_{0}^{\infty}\frac{1-e^{-t^{2}}}{t^{2}}dt=\sqrt{% \frac{n}{\sigma}}\sqrt{\pi}.$

Dividing (8) by $\sqrt{n}$ and taking the limsup then gives

 $\limsup_{n\to\infty}\frac{\pi}{2}\frac{I_{n}(|x|)}{\sqrt{n}}\leq\sqrt{\frac{% \pi}{2(1-\epsilon)}},$

or

 $\limsup_{n\to\infty}\frac{I_{n}(|x|)}{\sqrt{n}}\leq\sqrt{\frac{2}{\pi(1-% \epsilon)}};$

indeed $\beta_{\epsilon}$ depends on $n$, but $\beta_{\epsilon}<\frac{2}{\alpha_{\epsilon}}$, which does not depend on $n$. Taking $\epsilon\to 0$ yields

 $\limsup_{n\to\infty}\frac{I_{n}(|x|)}{\sqrt{n}}\leq\sqrt{\frac{2}{\pi}}.$

On the other hand, taking the liminf of (8) divided by $\sqrt{n}$ gives

 $\liminf_{n\to\infty}\frac{\pi}{2}\frac{I_{n}(|x|)}{\sqrt{n}}\geq\sqrt{\frac{% \pi}{2}},$

or

 $\liminf_{n\to\infty}\frac{I_{n}(|x|)}{\sqrt{n}}\geq\sqrt{\frac{2}{\pi}}.$

Combining the limsup and the liminf inequalities proves the claim. ∎

###### Lemma 8.

For any $\xi\in\mathbb{R}$ and $n\in\mathbb{N}$,

 $I_{n}(e^{\xi|x|})

We will use the following theorem to establish an estimate similar to but weaker than the law of the iterated logarithm.66 6 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 189, Solution 13.7.

###### Theorem 9.

For any $\epsilon>0$, for almost all $t\in[0,1]$,

 $\sum_{n=1}^{\infty}\frac{1}{n^{2+\epsilon}}\exp\left(\sqrt{\frac{2\log n}{n}}% \left|\sum_{k=1}^{n}r_{k}(t)\right|\right)dt<\infty.$
###### Proof.

Define $f_{n}:[0,1]\to(0,\infty)$ by

 $f_{n}(t)=\frac{1}{n^{2+\epsilon}}\exp\left(\sqrt{\frac{2\log n}{n}}\left|\sum_% {k=1}^{n}r_{k}(t)\right|\right).$

Applying Lemma 8 with $\xi=\sqrt{\frac{2\log n}{n}}$,

 $\int_{0}^{1}f_{n}(t)dt\leq\frac{1}{n^{2+\epsilon}}\cdot 2\cdot\left(\cosh\sqrt% {\frac{2\log n}{n}}\right)^{n}.$

It is not obvious, but we take as given the asymptotic expansion

 $\left(\cosh\sqrt{\frac{2\log n}{n}}\right)^{n}=n-\frac{1}{3}(\log n)^{2}+\frac% {\frac{8}{45}(\log n)^{3}+\frac{1}{18}(\log n)^{4}}{n}+O(n^{-3/2}),$

and using this,

 $\frac{1}{n^{2+\epsilon}}\cdot 2\cdot\left(\cosh\sqrt{\frac{2\log n}{n}}\right)% ^{n}=\frac{2}{n^{1+\epsilon}}+O\left(\frac{(\log n)^{2}}{n^{2+\epsilon}}\right% )=\frac{2}{n^{1+\epsilon}}+O(n^{-2}).$

Thus

 $\sum_{n=1}^{\infty}\int_{0}^{1}f_{n}(t)dt=\sum_{n=1}^{\infty}\left(\frac{2}{n^% {1+\epsilon}}+O(n^{-2})\right)<\infty.$

Because each $f_{n}$ is nonnegative, using this with the monotone convergence theorem gives the claim. ∎

###### Theorem 10.

For almost all $t\in[0,1]$,

 $\limsup_{n\to\infty}\frac{\left|\sum_{k=1}^{n}r_{k}(t)\right|}{\sqrt{n\log n}}% \leq\sqrt{2}.$
###### Proof.

Let $\epsilon>0$. By Theorem 9, for almost all $t\in[0,1]$ there is some $n_{t}$ such that $n\geq n_{t}$ implies that $f_{n}(t)<1$, where we are talking about the functions $f_{n}$ defined in the proof of that theorem; certainly the terms of a convergent series are eventually less than $1$. That is, for almost all $t\in[0,1]$ there is some $n_{t}$ such that $n\geq n_{t}$ implies that (taking logarithms),

 $(-2-\epsilon)\log n+\sqrt{\frac{2\log n}{n}}\left|\sum_{k=1}^{n}r_{k}(t)\right% |<0,$

and rearranging,

 $\frac{\left|\sum_{k=1}^{n}r_{k}(t)\right|}{\sqrt{n\log n}}<\sqrt{2}+\frac{% \epsilon}{\sqrt{2}}=\sqrt{2}+\epsilon^{\prime}.$

For each $s\in\mathbb{N}$, let $E_{s}$ be those $t\in[0,1]$ such that

 $\limsup_{n\to\infty}\frac{\left|\sum_{k=1}^{n}r_{k}(t)\right|}{\sqrt{n\log n}}% >\sqrt{2}+\frac{1}{s}.$

For each $s$, taking $0<\epsilon^{\prime}<\frac{1}{s}$ we get that almost all $t\in[0,1]$ do not belong to $E_{s}$. That is, for each $s$, the set $E_{s}$ has measure $0$. Therefore

 $E=\bigcup_{s=1}^{\infty}E_{s}$

has measure $0$. That is, for almost all $t\in[0,1]$, for all $s\in\mathbb{N}$ we have $t\not\in E_{s}$, i.e.

 $\limsup_{n\to\infty}\frac{\left|\sum_{k=1}^{n}r_{k}(t)\right|}{\sqrt{n\log n}}% \leq\sqrt{2}+\frac{1}{s},$

and this holding for all $s\in\mathbb{N}$ yields

 $\limsup_{n\to\infty}\frac{\left|\sum_{k=1}^{n}r_{k}(t)\right|}{\sqrt{n\log n}}% \leq\sqrt{2},$

completing the proof. ∎

## 3 Hypercubes

Let $m_{n}$ be Lebesgue measure on $\mathbb{R}^{n}$, and let $Q_{n}=[0,1]^{n}$.77 7 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 161, Solution 11.1.

###### Theorem 11.

If $f\in C([0,1])$, then

 $\lim_{n\to\infty}\int_{Q_{n}}f\left(\frac{x_{1}+\cdots+x_{n}}{n}\right)dm_{n}(% x)=f\left(\frac{1}{2}\right).$
###### Proof.

Define $X_{n}:Q_{n}\to\mathbb{R}$ by

 $X_{n}=\frac{x_{1}+\cdots+x_{n}}{n},\qquad x\in Q_{n}.$

We have

 $\int_{Q_{n}}X_{n}dm_{n}(x)=\frac{1}{n}\sum_{k=1}^{n}\int_{0}^{1}x_{k}\cdot 1dx% _{k}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2}=\frac{1}{2},$

and we define

 $\displaystyle V_{n}$ $\displaystyle=$ $\displaystyle\int_{Q_{n}}\left(X_{n}-\frac{1}{2}\right)^{2}dm_{n}(x)$ $\displaystyle=$ $\displaystyle\int_{Q_{n}}\sum_{k=1}^{n}\frac{x_{k}^{2}}{n^{2}}+\sum_{j\neq k}% \frac{x_{j}x_{k}}{n^{2}}-X_{n}+\frac{1}{4}dm_{n}(x)$ $\displaystyle=$ $\displaystyle\frac{1}{n^{2}}\sum_{k=1}^{n}\int_{0}^{1}x_{k}^{2}dx_{k}+\frac{1}% {n^{2}}\sum_{j=1}^{n}\sum_{k\neq j}\int_{0}^{1}x_{j}dx_{j}\int_{0}^{1}x_{k}dx_% {k}-\frac{1}{2}+\frac{1}{4}$ $\displaystyle=$ $\displaystyle\frac{1}{n^{2}}\sum_{k=1}^{n}\frac{1}{3}+\frac{1}{n^{2}}\sum_{j=1% }^{n}\sum_{k\neq j}\frac{1}{4}-\frac{1}{4}$ $\displaystyle=$ $\displaystyle\frac{1}{3n}+\frac{n-1}{4n}-\frac{1}{4}$ $\displaystyle=$ $\displaystyle\frac{n^{-1}}{12}.$

Suppose that $c_{n}$ is a sequence of positive real numbers tending to $0$, and define $J_{n}=J_{n}(c)$ to be those $x\in Q_{n}$ such that

 $\left|X_{n}(x)-\frac{1}{2}\right|\geq c_{n}.$

Then

 $\displaystyle V_{n}$ $\displaystyle=$ $\displaystyle\int_{Q_{n}}\left(X_{n}-\frac{1}{2}\right)^{2}dm_{n}(x)$ $\displaystyle\geq$ $\displaystyle\int_{J_{n}}\left(X_{n}-\frac{1}{2}\right)^{2}dm_{n}(x)$ $\displaystyle\geq$ $\displaystyle\int_{J_{n}}c_{n}^{2}dm_{n}(x)$ $\displaystyle=$ $\displaystyle c_{n}^{2}m_{n}(J_{n}),$

so

 $m_{n}(J_{n})\leq\frac{V_{n}}{c_{n}^{2}}=\frac{n^{-1}}{12c_{n}^{2}}.$

Take $c_{n}=n^{-1/3}$, giving

 $m_{n}(J_{n})\leq\frac{n^{-1/3}}{12}.$

Let $\epsilon>0$. Because $f$ is continuous, there is some $\delta>0$ such that $|t-\frac{1}{2}|<\delta$ implies that $|f(t)-f(\frac{1}{2})|<\epsilon$; furthermore, we take $\delta$ such that

 $\frac{\left\|f\right\|_{\infty}\delta}{6}<\epsilon.$

Set $N>\delta^{-3}$. For $n\geq N$ and $x\in Q_{n}\setminus J_{n}$,

 $\left|X_{n}(x)-\frac{1}{2}\right|

and so

 $\left|(f(X_{n}(x))-f\left(\frac{1}{2}\right)\right|<\epsilon.$

This gives us

 $\displaystyle\left|\int_{Q_{n}}f(X_{n}(x))dm_{n}(x)-f\left(\frac{1}{2}\right)\right|$ $\displaystyle=$ $\displaystyle\left|\int_{Q_{n}}f(X_{n}(x))-f\left(\frac{1}{2}\right)dm_{n}(x)\right|$ $\displaystyle\leq$ $\displaystyle\int_{J_{n}}\left|f(X_{n}(x))-f\left(\frac{1}{2}\right)\right|dm_% {n}(x)$ $\displaystyle+\int_{Q_{n}\setminus J_{n}}\left|f(X_{n}(x))-f\left(\frac{1}{2}% \right)\right|dm_{n}(x)$ $\displaystyle\leq$ $\displaystyle\int_{J_{n}}2\left\|f\right\|_{\infty}dm_{n}(x)+\int_{Q_{n}% \setminus J_{n}}\epsilon dm_{n}(x)$ $\displaystyle\leq$ $\displaystyle 2\left\|f\right\|_{\infty}m_{n}(J_{n})+\epsilon$ $\displaystyle\leq$ $\displaystyle 2\left\|f\right\|_{\infty}\frac{n^{-1/3}}{12}+\epsilon$ $\displaystyle<$ $\displaystyle\frac{\left\|f\right\|_{\infty}\delta}{6}+\epsilon$ $\displaystyle<$ $\displaystyle 2\epsilon,$

which proves the claim. ∎