Relation between function discontinuities and Fourier transform at infinity
Proposition. Suppose that $f$ is $(m-1)$ times continuously differentiable, and let $g=f^{(m-1)}$. Suppose that $g$ tends to $0$ at infinity, and there exists a finite nonempty set $D$ such that
- $g'$ exists on $\mathbb R\setminus D$
- There exists $h\in L^1(\mathbb R)$ such that $g'(x)-\int_{-\infty}^xh(t)\,dt$ is constant on each connected component of $\mathbb R\setminus D$.
Then $|\xi|^{m+1}|\hat f(\xi)|$ is bounded as $\xi\to\infty$, but does not tend to zero.
Proof. Since $|\hat f(\xi)|$ is equal to $|\xi|^{m-1}|\hat g(\xi)|$ up to some constant factor, it suffices to work with $ |\xi|^{2}|\hat g(\xi)|$. Split the integral defining $\hat g$ into integrals over connected components of $\mathbb R\setminus D$, denoted $(a_k,a_{k+1})$ below, $-\infty=a_0<\dots<a_n=+\infty$. And integrate by parts: $$ \hat g(\xi)=\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} g(x)\,dx = \frac{1}{i\xi}\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} g'(x)\,dx \tag1 $$ No boundary terms appear because $g$ is continuous and vanishes at infinity. But they do appear when we integrate again, turning (1) into $$ \frac{-1}{\xi^2}\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} h(x)\,dx -\frac{1}{\xi^2} \sum_{k=1}^{n-1} e^{-i\xi a_k} (g'(a_k-)-g'(a_k+)) \tag2 $$ By the Riemann-Lebesgue lemma, the first integral in (2) tends to zero as $\xi\to\infty$. Therefore, $|\xi|^2|\hat g(\xi)|=o(1)+|P(\xi)|$, where $P$ is a nonzero trigonometric polynomial. $\Box$
Remarks
- In the special case when $D$ consists of one point, $|P|$ is constant and therefore $|\xi|^{m+1}|\hat f(\xi)|$ has a finite nonzero limit at infinity.
- Condition 2 can be expressed by saying that $g'$ is absolutely continuous on $\mathbb R\setminus D$ with integrable derivative.