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I have been proving some facts about Rademacher functions:

I know that the Rademacher functions are: $r_n(t)=\operatorname{sign}(\sin{2^n\pi t})$

or, equivalently: $$r_m(t):=\left\{\begin{array}{cl}-1 & \mbox{if }t\in\displaystyle\bigcup_{k=1}^{2^{m-1}} \left(\frac{2k-1}{2^m},\frac{2k}{2^m}\right)\\&\\0,&\mbox{if }t\in\left\{\displaystyle\frac{k}{2^m}\,:\,k=1,2,\ldots,2^m\right\}\\&\\1,&\mbox{if }t\in\displaystyle\bigcup_{k=1}^{2^{m-1}}\left(\frac{2k-2}{2^m},\frac{2k-1}{2^m}\right)\end{array}\right.$$

I know the next properties:

$\triangleright$ If $n>m$ then $\displaystyle\int_{I_j^m} r_n=0$, where $I_j^m=\left[\displaystyle\frac{j}{2^m},\frac{j+1}{2^m}\right],0\leq j<2^m$.

$\triangleright$ If $n_1<n_2<\ldots<n_k$, then $\displaystyle\int r_{n_1}\ldots r_{n_k}=0$

$\triangleright$ $(r_n)_{n\geq 0}$ is an orthonormal system in $L_2([0,1])$ that is not basis.

I have just proved this inequality: $$\displaystyle\int_0^1\left|\displaystyle\sum_{i=1}^n a_i r_i(t)\right|^4\leq 3\left(\displaystyle\sum_{i=1}^n |a_i|^2\right)^2$$ where $a_1,\ldots, a_n$ are scalars (real or complex).

Now I need to show that the sequence $\left(\displaystyle\sum_{i=1}^n \frac{r_i(t)}{n}\right)$ converges t0 $0$ almost everywhere in $[0,1]$ using the last inequality (is assumed that is immediate, but I don't control the convergence a.e.). I thought to use $a_i=\frac{1}{n}$ or something similar, but I don't know.

Many thx.

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Let $a_1, \ldots, a_n = n^{-1}$ then we have

$$\int_0^1 \left |\sum_{i = 1}^n \frac{r_i(t)}{n} \right |^4 \, \textrm{d}t \leqslant \frac{3}{n^2}.$$

Hence taking the limit $n \to \infty$ we obtain $$\int_0^1 \left |\sum_{i = 1}^n \frac{r_i(t)}{n} \right |^4 \, \textrm{d}t \to 0.$$

Now we can apply Fatou's lemma to deduce that

$$\int_0^1 \lim_{n \to \infty} \left| \sum_{i = 1}^n \frac{r_i(t)}{n} \right|^4 \, \textrm{d}t = 0.$$

Hence

$$\lim_{n \to \infty} \left| \sum_{i = 1}^n \frac{r_i(t)}{n} \right|^4 = 0 \text{ a.e..}$$

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    $\begingroup$ @ Jonas Teuwen: For $f_n(t)=|n^{-1}\sum_{i=1}^n r_i(t)|^4$, Fatou's lemma only gives $\int_0^1\liminf_{\,n\to\infty}f_n(t){\kern.4mm\rm d\kern.4mm}t = 0$ from which you can conclude that $\liminf_{\,n\to\infty}f_n = 0$ almost everywhere. Wherefrom do you conclude that $\liminf_{\,n\to\infty}f_n(t)< \limsup_{\,n\to\infty}f_n(t)$ holds only for $t$ in a set of measure zero? $\endgroup$ – The-unKnowN Sep 22 '13 at 15:03

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