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$L^p$ to it meansits 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