$L^1$ norm for a product of cosines

Question

Let $k$ be an integer and consider the function $$ f(t)=\prod_{i=1}^{k} \cos(3^{i-1}\pi t). $$ I'm interested in finding bounds for $\int_{0}^{1}|f(t)|dt$ in terms of $k$. The first idea that comes to mind is using Hölder's inequality: $$ \int_{0}^{1} |f(t)| dt \le \prod_{i=1}^{k} \bigg(\int_{0}^{1}|\cos(3^{i-1}\pi t)|^k dt \bigg)^{\frac{1}{k}} $$ Now, using the periodicity of $|\cos(3^{i-1}\pi t)|$, note that $$ \int_{0}^{1}|\cos(3^{i-1}\pi t)|^k dt =\int_{0}^{1}|\cos(\pi t)|^k dt, $$ so that $$ \int_{0}^{1} |f(t)|dt \le \int_{0}^{1}|\cos(\pi t)|^k dt, $$ and I'm sure there should be some results for the moments of cosine. However, computationally this doesn't seem like a good bound. Is there a better way to do this? The reason I'm interested in this function is because it appears in the context of the Fourier transform of a certain set of integers.

Answer 1

Since I was asked to in the comments, let me record here some very simple observations. They boil down to the following inequality: for all $k \ge N$ we have

$$I_{k - N} \min_{x\in \mathbb{R}} h_N(x) \le I_k \le I_{k-N} \max_{x\in \mathbb{R}} h_N(x),$$ where $I_k$ is the integral $\int_0^1 |f_k(t)|dt$ from the OP (I write subscript $k$ to avoid confusion), and $h_N(x) = \frac{1}{3^N}\sum_{j = 0}^{3^N-1}\left|f_N\left(x + \frac{j}{3^N}\right)\right|$.

Indeed, by the change of variables and/or periodicity we have

$$I_{k-N} = \int_0^1 |f_{k-N}(t)|dt = \int_0^1 |f_{k-N}(3^Nt)|dt.$$

To get $f_k(t)$ from $f_{k-N}(3^Nt)$ we have to multiply by the first $N$ cosine factors. But the idea is that if we replace $t$ with $t + \frac{j}{3^N}$ the value of $f_{k-N}(3^Nt)$ doesn't change. So, we can add all this points (considered modulo $1$) together and get $$I_k = \int_0^1 h_N(t)|f_{k-N}(3^Nt)|dt,$$ from which we get the bounds I claimed in the beginning.

Applying this for $N = 1$ (and using deus ex machina aka Wolfram Alpha to find the minimal and maximal value) we get $$\frac{\sqrt{3}}{3}I_{k-1} \le I_k \le \frac{2}{3} I_{k-1},$$ so $\left(\frac{\sqrt{3}}{3}\right)^k \le I_k \le \left(\frac{2}{3}\right)^k$. Note that the upper bound is already better than the one from my earlier comment with $L^1-L^2$ norm estimate since $\frac{2}{3} < \frac{\sqrt{2}}{2}$. Using $N = 2$ provably gives even better bounds but only marginaly and I can't compute them explicitly.

It is reasonable to conjecture that, as we slowly tend $N$ to infinity, in this way we can get the optimal exponent, that is $$\lim_{N\to \infty} \left(\min_{x\in \mathbb{R}} h_N(x)\right)^{1/N} = \lim_{N\to \infty} \left(\max_{x\in \mathbb{R}} h_N(x)\right)^{1/N} = c,$$ but I don't know how to prove it and what the value of $c$ should be (at the risk of being once again wrong let me conjecture that $c = \int_0^1 |\cos(\pi t)|dt = \frac{2}{\pi}$).

Answer 2

Another idea (from the work of Maynard on primes with missing digits) to bound this integral is as follows: Since $t\in [0,1]$, we expand $t$ in base 3 as $t=\sum_{i=1}^{\infty} t_i/3^i$, where $t_i\in \{0,1,2\}$. Now, note that $$ 3^{i-1}t= \mbox{Integer} + \frac{t_i}{3}+\frac{t_{i+1}}{3^2}+u_i, $$ where $u_i \in [0,1/3^2]$. Therefore, since $|\cos(3^{i-1}\pi t)|$ is periodic with period $1$, then $$ f(t) \le \prod_{i=1}^{k-1} G(t_i,t_{i+1}), $$ where $$ G(a,b)=\sup_{u\in[0,1/3^2]}\bigg|\cos\big(\pi\big(\frac{a}{3}+\frac{b}{3^2}+u\big)\big)\bigg|. $$ Now, consider the $3\times 3$ matrix defined at the $i,j$ entry by $M[i,j]=G(i-1,j-1)$ for $i,j=1,2,3$. Then, it is possible to show via the Perron Frobenius theorem (see https://arxiv.org/abs/1604.01041 for details) that $$ \int_{0}^{1}|f(t)|dt \ll \big(\frac{\lambda}{3}\big)^k, $$ where $\lambda$ is the largest eigenvalue of the matrix. Intuitively, we are approximating $f(t)$ by the value at the first two digits that matter in the expansion. A numerical computation reveals that $\lambda\approx 2.19429$. If instead of approximating by the first two decimals we instead approximate by the first 5 decimals, say $$ 3^{i-1}t= \mbox{Integer} + \frac{t_i}{3}+\frac{t_{i+1}}{3^2}+\frac{t_{i+2}}{3^3}+\frac{t_{i+3}}{3^4}+\frac{t_{i+4}}{3^5}+u_i, $$ where $u_i\in[0,1/3^5]$ and then we consider certain $3^4\times 3^4$ matrix, then it's possible to show that $$ \int_{0}^{1}|f(t)|dt \ll \big(\frac{1.94884}{3}\big)^k, $$ and again the 1.94884 comes from the maximal eigenvalue of that matrix. It would be interesting to consider the sequence of eigenvalues and conjecture what the limit is. For 10 digits, the value is approximately 1.94495.