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
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 $\xi$.