# 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.

## 1 comment

1. Jacky Wong says:

不明觉厉。