Categories

## Alternating Fourier Coefficients

Suppose $f\in L^2$ is a periodic function from $\mathbb{R}$ to $\mathbb{C}$ with period $1$. Let $a(n)$ be its Fourier coefficients, namely $a(n)=\int_0^1{f(x)e^{-i2\pi nx}dx}$ for all $n\in\mathbb{Z}$. Prove for all $p\geq 2$ it is almost surely that function $g(\omega)(x)=\sum_{n\in \mathbb{Z}}\omega(n)a(n)e^{i2\pi nx}$ is in $L^p$ where $\omega$ is an infinite sequence of independent and identical random variables indexed by $\mathbb{Z}$ with $\omega(n)$ equals either $-1$ or $1$ with probability $1/2$.

I heard this problem from Yue Pu who heard it from Gautam Iyer. Roughly speaking, given a periodic $L^2$ function, by alternating the signs of its Fourier coefficients, one almost surely gets an $L^p$ function for all $p\geq 2$.

Here is the proof Yue and I worked out this afternoon which I think is pretty neat. However, all convergence arguments are omitted.

We shall prove that for all $p\in\mathbb{N}$, the expectation of $\int_0^1 |g(\omega)(x)|^{2p}dx$ is less than or equal to $C_p\left(\int_0^1|f(x)|^2dx\right)^p$, where $C_p$ is a constant that only depends on $p$.

This is sufficient to prove the original problem. For all $p\in\mathbb{N}$, we let $A=\{\omega: g(\omega)\notin L^{2p}\}$. If $\mathbb{P}(A)>0$, then for all $\omega\in A$, $\int_0^1{|g(\omega)(x)|^{2p}dx}=\infty$. Consequently $$\infty\leq\mathbb{E}\left[\int_0^1 |g(\omega)(x)|^{2p}dx\right]\leq C_n\left(\int_0^1|f(x)|^2dx\right)^p<\infty,$$ which is impossible. For a general $p\geq 2$, pick any $p'\in\mathbb{N}$ such that $p\leq 2p'$. We know $g(\omega)$ is almost surely in $L^{2p'}$, hence almost surely in $L^{p}$.

For brevity, we denote $e(n)(x)=e^{i2\pi nx}$. We will repeatedly use the facts that $e(m)(x)e(n)(x)=e(m+n)(x)$, $\overline{e(m)(x)}=e(-m)(x)$ and $\int_0^1e(n)(x)=0$ except for $n=0$ it equals $1$. We will also use Parseval’s identity which says $\int_0^1|f(x)|^2dx=\sum_{n\in\mathbb{Z}}|a(n)|^2$.

For all $p\in\mathbb{N}$, we have \begin{aligned}|g(\omega)(x)|^2&=g(\omega)(x)\overline{g(\omega)(x)}\\&=\sum_{m,n}\omega(m)\omega(n)a(m)\overline{a(n)}e(m)(x)e(-n)(x).\end{aligned} This gives us $$|g(\omega)(x)|^{2p}=\sum_{m_1,\ldots,m_p,n_1,\ldots,n_p}\prod_{i=1}^p\omega(m_i)a(m_i)e(m_i)(x)\omega(n_i)\overline{a(n_i)}e(-n_i)(x).$$ Integrate above formula from $0$ to $1$, we have
$$\int_0^1{|g(\omega)(x)|^{2p}}dx=\sum_{m_1+\ldots+m_p=n_1+\ldots+n_p}\prod_{i=1}^p\omega(m_i)\omega(n_i)a(m_i)\overline{a(n_i)}.$$

Notice that for fixed $m_1, \ldots, m_p, n_1, \ldots, n_p$, the expectation of the summand is always $0$ except when there is a perfect matching among $m_1,\ldots, m_p, n_1, \ldots, n_p$. By a perfect matching, we mean that one can pair the numbers such that the two numbers in each pair are the same. Hence the expectation of above is equal to
$$\sum_{m_1+\ldots+m_p=n_1+\ldots+n_p}^*\prod_{i=1}^pa(m_i)\overline{a(n_i)},$$
where $\sum^*$ is the summation over every sequence that admits a perfect matching.

The above is less than or equal to
$$\sum_{m_1,\ldots, m_p, n_1,\ldots, n_p}^*\prod_{i=1}^p|a(m_i)||a(n_i)|.$$

By over counting, the expectation is less than or equal to
$$\frac{(2p)!}{p!2^p}\sum_{l_1, \ldots, l_p}\prod_{i=1}^p|a(l_i)|^2=\frac{(2p)!}{p!2^p}\left(\int_0^1|f(x)|^2dx\right)^p,$$
where $\frac{(2p)!}{p!2^p}$ is the number of perfect matching can possibly be formed among $2p$ numbers.

Remark. The proof is not constructive. Indeed, it doesn’t provide a way for one to alternate the Fourier coefficients so that the outcome is an $L^p$ function, though it is almost sure that the outcome is in $L^p$.