$\textbf{Lemma}$ If $f\in L^1(\mathbb{R}^d)$ and \begin{align*} \hat{f}(\xi)=\int_{\mathbb{R}^d} f(x)e^{-2\pi i x \xi}dx, \end{align*} then $\hat{f}(\xi) \rightarrow 0 $ as $\vert \xi \vert \rightarrow \infty$.
$\textbf{Problem} $ Let $f \in L^1$. Define \begin{align*} \hat f (t) := \int_{-\infty}^{\infty} f(x) e^{-2\pi i x t}dx \end{align*} and for $u>0$ \begin{align*} S_u(x) := \int_{-u}^u \hat f(t) e^{2\pi i t x} dx \end{align*} Suppose that for given $x_0\in \mathbb{R}$, the following holds: \begin{align*} \int_0^{\infty} \vert \frac{f(x_0+y)+f(x_0-y)-2f(x_0)}{y}\vert dy <\infty . \end{align*} Use Riemann - Lebesgue Lemma to show that \begin{align*} \lim_{u \rightarrow \infty} S_u(x_0) = f(x_0) \end{align*} (Hint: You can use $\int_0^{\infty} \frac{\sin t } {t} dt = \pi/2$)
My Attemt: If $D_u(x)=\int_{-u}^u e^{2\pi i tx}dt$, I got \begin{align*} D_u(x)= \frac{\sin (2\pi ux)}{\pi x} , \quad S_u(x)=f*D_u(x). \end{align*} (where * operation means convolution)
Then, \begin{align*} S_u(x_0) &= f*D_u(x_0)\\ &=\int_{-\infty}^{\infty} f(x_0-y) D_u(y) dy \\ &=\int_{-\infty}^{\infty} f(x_0-y) \frac{\sin (2\pi u y)}{\pi y} dy \\ &=\int_{-\infty}^{\infty} f(x_0+y) \frac{\sin (2\pi u y)}{\pi y} dy \\ \end{align*} Consequently, $S_u(x_0) = \frac{1}{\pi} \int_0^{\infty} \frac{f(x_0-y)+f(x_0+y)}{y} \sin(2\pi u y)dy$ and $f(x_0)=\frac{2}{\pi}\int_0^\infty \frac{\sin( 2\pi u y)}{2\pi u y}f(x_0)dy$.
However, $\vert S_u(x_0) - f(x_0) \vert \leq \frac{1}{\pi} \int_0^\infty \vert(\frac{f(x_0+y)+f(x_0-y)}{y} - \frac{2f(x_0)}{2\pi u y} )(\sin (2\pi u y))\vert dy$
How to get the conclusion by using the Riemann-Lebesgue Lemma?
Any help is appreciated...
Thank you!
Since $\displaystyle \int_0^\infty \frac{\sin 2\pi uy}{y}\, dy = \frac{\pi}{2}$, then $$f(x_0) = \int_0^\infty \frac{\sin 2\pi uy}{y} 2f(x_0)\, dy$$ Thus $$f(x_0) - S_u(x_0) = -\frac{1}{\pi}\int_0^\infty \sin(2\pi uy)\, \frac{f(x_0 + y) + f(x_0 - y) - 2f(x_0)}{y}\, dy\tag{*}\label{eq1}$$ Let $h$ be the function on $\Bbb R$ that is equal to $\dfrac{1}{\pi}\dfrac{f(x_0 + y) + f(x_0 - y) - 2f(x_0)}{y}$ for $y>0$ and equal to $0$ elsewhere. The integral condition in the problem statement implies $h\in L^1(\Bbb R)$. By the Riemanmn-Lebesgue lemma, the imaginary part of $\hat{h}(u)$ tends to $0$ as $u \to \infty$. But the imaginary part of $\hat{h}(u)$ is the integral in \eqref{eq1}. The result now follows.