Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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

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

share|improve this answer
    
@ 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? –  The-unKnowN Sep 22 '13 at 15:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.