Let $u\in L^1(\mathbb{S}^{d-1})$. I want to show that
\begin{align*}
\lim_{|\xi|\to \infty}
\int_{\mathbb{S}^{d-1}}(1-\cos(\xi\cdot w))u(w)d \sigma_{d-1}(w)
= \int_{\mathbb{S}^{d-1}}u(w)d \sigma_{d-1}(w).
\end{align*}
Basically, the question can be reduced into showing that
\begin{align*}
\lim_{|\xi|\to \infty}
\int_{\mathbb{S}^{d-1}}\cos(\xi\cdot w)u(w)d \sigma_{d-1}(w)
= 0.
\end{align*}
This looks like a Riemann-Lebesgue lemma. But I don't know how to tackle it.
I intuitively guessed this from the classical Riemann-Lebesgue Lemma which infers that
\begin{align*}
\lim_{|\xi|\to \infty}
\int_{B_1(0)}(1-\cos(|\xi| z\cdot x))u(x)d x
= \int_{B_1(0)}u(x)dx\quad \text{for fixed $z\in \Bbb R^d$}.
\end{align*}
More generally, if $f$ is $T^d$-periodic, then $f_\lambda(x)= f(\lambda x)$ weakly converge in $L^p$ to its mean value as $\lambda\to\infty$ that is
$$f_\lambda \rightharpoonup \bar f,\quad \quad \bar f=\frac{1}{T^d}\int_{[0,T]^d}f(x) dx.$$
Is there any good reference for this type of limit?
Any help is welcome
Best Answer
Throughout the rest of the posting, the direction $\mathbf{v}=|\boldsymbol{\xi}|^{-1}\boldsymbol{\xi}$ is fixed. Let $R_v$ be a unitary transformation of $\mathbb{R}^d$ so that $R^\intercal \big(\mathbf{v})=\mathbf{e}_d$. Then \begin{align} \int_{\mathbb{S}_d}\big(1-\cos(\boldsymbol{\xi}\cdot \mathbf{w})\big)&u(w) \sigma_{n-1}(dw)=\int_{\mathbb{S}_d}\big(1-\cos(|\boldsymbol{\xi}|w_d)\big)u(R_vw)\sigma_{n-1}(d\mathbf{w})\\ &=\int^1_{-1}(1-\cos(|\boldsymbol{\xi}|w_d))(1-|w_d|^2)^{\frac{d-3}{2}}\Big(\int_{\mathbb{S}_{d-2}}u(R_v(\mathbf{z},w_d))\,\sigma_{d-2}(d\mathbf{z})\Big)\,dw_d\\ &=\int^1_{-1}(1-\cos(|\boldsymbol{\xi}|w_d))G_v(w_1)\,dw_1\xrightarrow{|\boldsymbol{\xi}|\rightarrow\infty}\int^1_{-1}G_v(w_1)\,dw_1\\ &=\int_{\mathbb{S}_{d-1}}u(R_v\mathbf{w})\,\sigma_{d-1}(d\mathbf{w})=\int_{\mathbb{S}_{d-1}}u(\mathbf{w})\,\sigma_{d-1}(d\mathbf{w}) \end{align} In the third line we apply Riemann-Lebesgue's Lemma (or Féjer's Lemma). Notice that the limit is mean to be taking along the ray $\{av:a>0\}$.
Edit: This I think the directional limit is all one can obtain through direct spherical coordinate methods. To get a general results that allows to vary the direction of $\xi$ a little more Fourier analysis is required.
Denote by $\lambda_d$ the Lebesgue measure on $\mathbb{R}^d$. Recall the following results from Fourier analysis
Given a finite Radon measure $\mu$ on $\mathbb{R}^d$, its Fourier transform $\widehat{\mu}$ is defined as $$\widehat{\mu}(\mathbf{t})=\int e^{-2\pi i\mathbf{t}\cdot \mathbf{x}}\,\mu(d\mathbf{x})$$ It is easy to check that $\widehat{\mu}$ is uniformly continuous bounded function.
Proof: By Fubini-Tonelli's theorem \begin{align} \widehat{f\cdot\mu}(\mathbf{t})&=\int e^{-2\pi i\mathbf{t\cdot y}}f(\mathbf{y})\mu(d\mathbf{y})=\int e^{-2\pi i\mathbf{t\cdot y}}\Big(\int e^{2\pi i \mathbf{s\cdot y}}\widehat{f}(\mathbf{s})\,d\mathbf{s}\Big)\mu(d\mathbf{y})\\ &=\int \widehat{f}(\mathbf{s})\Big(\int e^{-2\pi i(\mathbf{t-s})\cdot y}\,\mu(d\mathbf{y})\Big)\,d\mathbf{t}\\ &=\int \widehat{f}(\mathbf{s})\widehat{\mu}(\mathbf{t-s})\,d\mathbf{s}=(\widehat{f}*\widehat{\mu})(\mathbf{t}) \end{align}
To go back to the OP, consider the measure $\sigma_{d-1}$ as a measure on $\mathbb{R}^d$ supported on the sphere $\mathbb{S}_{d-1}$.
Functions in $L_1(\sigma_{d-1})$ can be approximated by smooth functions. Suppose first that $u\in C^{\infty}(\mathbb{S}_{d-1})$. Let $\rho\in C^{\infty}(\mathbb{R})$ such that $0\leq \rho\leq 1$, $\rho(1)=1$ and $\operatorname{supp}(\rho)=[1/2,3/2]$. The function $U(\mathbf{x})=\rho(|\mathbf{x}|) u(|\mathbf{x}|^{-1}\mathbf{x})$ for $\mathbf{x}\neq\boldsymbol{0}$ and $U(\boldsymbol{0})=0$ is a smooth function on $\mathbb{R}^d$ of compact support. Moreover, $u\cdot \sigma_{d-1}=U\cdot \sigma_{d-1}$ since $U=u$ on $\mathbb{S}_{d-1}$. Consequently $$\widehat{u\cdot \sigma_{d-1}}(t)=\widehat{U}*\widehat{\sigma_{d-1}}$$ The choice of $U$ implies that $\widehat{U}$ is a Schwartz function and so, $\widehat{U}\in L_p(\lambda)$ for all $p\geq1$.
This in particular, for $d=1$, Fact 3 implies that $\phi_m(r)=r^{-m} J_m(r)\xrightarrow{r\rightarrow\infty}0$ for all $m>-1/2$. Hence $\widehat{\sigma_{d-1}}=\phi_{\frac{d-2}{2}}$ decays to $0$ for each dimension $d\geq2$.
The conclusion then boils down to checking whether $\widehat{U}*\widehat{\sigma_{d-1}}(\mathbf{t})$ decays to $0$ as $|\mathbf{\mathbf{t}}|\rightarrow0$. One this has been established, the conclusion for all $u\in L_1(\sigma_{d-1})$ follows for density arguments.
$\widehat{\sigma_{d-1}}$ is bounded and so $\int_{B_d(0;1)}|\widehat{\sigma_{d-1}}|^p\,d\lambda_d<\infty$ for all $p\geq1$. Outside $B(0;1)$ we have \begin{align}\int_{B(0;1)^c}|\widehat{\phi_{d-1}}(\mathbf{t)}|^p\,d\mathbf{t}&\leq K_d\int^\infty_1r^{-(d-1)(\frac{p}{2}-1)}\,dr<\infty \end{align} whenever $(d-1)(\frac{p}{2}-1)>1$, that is $p>2+\frac{2}{d-1}$. This means that $\widehat{\sigma_{d-1}}\in L_p(\lambda_d)$ for all $p>2+\frac{2}{d-1}$. As $\widehat{U}\in L_{p'}(\lambda_d)$, where $\frac{1}{p}+\frac{1}{p'}=1$, it follows that $\widehat{U}*\widehat{\sigma_{d-1}}\in C_0(\mathbb{R}^d)$.
Comments:
Facts 2, 3 and 4 are well known results in Classic Fourier Analysis.
Fact 2 is a simple computation using spherical coordinates.
Facts 3 and 4 can be found for example in Jones, F., Lebesgue Measure on Euclidean space, Jones and Bartlett,2001, pp 344-347, or in Grafakos, L., Classical Fourier Analysis, 3rd Edition, Springer, 2014, pp. 577-580.
As for the statement at the end of the OP's posting, the statement holds for bounded measurable functions on $\mathbb{R}^d$ which are $T$-periodic in each component. However, notice that periodization yields integration over the $d$=dimensional torus $\mathbb{T}^d=(\mathbb{S}_1)^d$, not over the sphere $\mathbb{S}_{d-1}$ in $\mathbb{R}^d$. That can be shown by considering first a simple step function $\mathbb{1}_{[\boldsymbol{a},\boldsymbol{b}]}$ where $[\boldsymbol{a},\boldsymbol{b}]=[a_1,b_1]\times\ldots\times[a_d,b_d]$.