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

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
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.

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