Computing a limit on the unit sphere: Riemann Lebesgue?

Question

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

Answer 1

The key fact here is the (surprising, initially, but well known) (power) decay of $\widehat{\sigma}(\xi)$. If $u\in C^{\infty}(S)$, we can extend to a function $u_0\in C^{\infty}_0(\mathbb R^d)$, and then $$ \widehat{u\,d\sigma}=\widehat{u_0\,d\sigma}=\widehat{u_0}*\widehat{\sigma} $$ still decays. See here for the general version of the convolution theorem needed here.

We can then extend this to arbitrary $u\in L^1$ by the argument from the traditional Riemann-Lebesgue lemma: given $\epsilon>0$, pick a $v\in C^{\infty}(S)$ with $\|u-v\|_1<\epsilon$. Since $|\widehat{u\, d\sigma}(\xi) -\widehat{v\, d\sigma}(\xi)|<\epsilon$ and $\widehat{v\, d\sigma}\to 0$, we also have $|\widehat{u\, d\sigma}(\xi)|<2\epsilon$ for all large $\xi$.

Answer 2

Here are some useful and well facts from Classical Fourier Analysis that may be used to solve this problem.

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.

Fact 1: Assume $f\in C_0(\mathbb{R}^d)\cap L_1(\lambda)$ with $\widehat{f}\in L_1(\lambda)$,and that $\mu$ is a finite Radon measure on $\mathbb{R}^d$. Then, $f\cdot\mu$ is a well defined Radon measure on $\mathbb{R}^d$, and its Fourier transform is given by $\widehat{f}*\widehat{\mu}$.

This follows by direct application of Fubini's theorem.

To go back to the OP's problem. Consider $\sigma_{d-1}$ as a measure on $\mathbb{R}^d$ supported on the sphere $\mathbb{S}^{d-1}$.

Fact 2: The Fourier transform $\widehat{\sigma_{d-1}}(\mathbf{t})$ is known to be a radial function given by \begin{align} \widehat{\sigma_{d-1}}(\mathbf{t})=c_d|\mathbf{t}|^{-\frac{d-2}{2}} J_{\frac{d-2}{2}}(2\pi |\mathbf{t}|) \end{align} where $J_m$ ($m>-1$)is the Bessel function of first kind, and $c_d$ is a constant depending on dimension $d$.

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 $p\geq1$.

Fact 3: For any $a>-1$, $f_a(x)=(1-|x|^2)^a_+$ is in $L_1(\lambda_d)$ and \begin{align} \widehat{f_a}(\mathbf{t})=k_{a, d}|\mathbf{t}|^{-\frac{d}{2}-a}J_{\frac{d}{2}+a}(2\pi |\mathbf{t}|) \end{align} for some constant $k_{a, d}$ depending on $a$ and the dimension $d$.

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 $d\geq2$.

As it was pointed out in Christian Remling's solution, the 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.

Fact 4: It is known that $|J_m(t)|\leq C(m) t^{-1/2}$ for some $m>-\frac12$ and $t\geq 1$, where $C(m)$ is a constant depending on $m$

$\widehat{\sigma_{d-1}}$ 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 $p>2+\frac{1}{d-2}$. 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 $\widehat{U}*\widehat{\sigma_{d-1}}\in C_0(\mathbb{R}^d)$.

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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 $[\boldsymbol{a},\boldsymbol{b}]=[a_1,b_1]\times\ldots\times[a_d,b_d]$.