For $a > 0$ and $n \in \mathbb Z_+$, consider the oscillatory integral

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx,$$

where $f$ is an integrable function on $[0, 1]$, which we extend by periodicity to all of $\mathbb R$.

By a Riemann-Lebesgue type argument, one can show that we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx \to \left ( \int f(x) \, dx \right )^{n+1}$$

as $a \to \infty$. In particular the limit is $0$ for mean zero functions. We can see this as a “zeroth order” result, in the sense that it says the integral is $o(1)$ as $a \to \infty$.

Question: Assume now additional smoothness on $f$ such as Lipschitz or $C^k$. Can we obtain sharper asymptotics on the integral for mean zero functions?

For instance, do we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx = O(\frac{1}{a^n})$$

as $a \to \infty$, with the implied constants depending only on some Lipschitz or $C^k$ norm of $f$?

For $a > 0$ and $n \in \mathbb Z_+$, consider the oscillatory integral

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx,$$

where $f$ is an integrable function on $[0, 1]$, which we extend by periodicity to all of $\mathbb R$.

By a Riemann-Lebesgue type argument, one can show that we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx \to \left ( \int_0^1 f(x) \, dx \right )^{n+1}$$

as $a \to \infty$. In particular the limit is $0$ for mean zero functions. We can see this as a “zeroth order” result, in the sense that it says the integral is $o(1)$ as $a \to \infty$.

Question: Assume now additional smoothness on $f$ such as Lipschitz or $C^k$. Can we obtain sharper asymptotics on the integral for mean zero functions?

For instance, do we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx = O\left(\frac{1}{a^n}\right)$$

as $a \to \infty$, with the implied constants depending only on some Lipschitz or $C^k$ norm of $f$?

For $a > 0$ and $n \in \mathbb Z_+$, consider the oscillatory integral

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx,$$

where $f$ is an integrable function on $[0, 1]$, which we extend by periodicity to all of $\mathbb R$.

By a Riemann-Lebesgue type argument, one can show that we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx \to \left ( \int f(x) \, dx \right )^{n+1}$$

as $a \to \infty$. In particular the limit is $0$ for mean zero functions. We can see this as a “zeroth order” result, in the sense that it says the integral is $o(1)$ as $a \to \infty$.

Assume now additional smoothness on $f$ such as Lipschitz or $C^1$.

Question: Can we obtain sharper asymptotics on the integral for mean zero functions? For instance, do we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx = O(\frac{1}{a^n})$$

as $a \to \infty$, with the implied constants depending only on some Lipschitz or $C^k$ norm of $f$?

For $a > 0$ and $n \in \mathbb Z_+$, consider the oscillatory integral

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx,$$

where $f$ is an integrable function on $[0, 1]$, which we extend by periodicity to all of $\mathbb R$.

By a Riemann-Lebesgue type argument, one can show that we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx \to \left ( \int f(x) \, dx \right )^{n+1}$$

as $a \to \infty$. In particular the limit is $0$ for mean zero functions. We can see this as a “zeroth order” result, in the sense that it says the integral is $o(1)$ as $a \to \infty$.

Question: Assume now additional smoothness on $f$ such as Lipschitz or $C^k$. Can we obtain sharper asymptotics on the integral for mean zero functions?

For instance, do we have

$$\int_{0}^1 f(x) f(ax) \dots f(a^n x) \, dx = O(\frac{1}{a^n})$$

as $a \to \infty$, with the implied constants depending only on some Lipschitz or $C^k$ norm of $f$?