- Prove the Riemann-Lebesgue Lemma: if \(f \in L^1(\mathbb{R}^d)\), then the Fourier transform of \(f\),
\[\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x) e^{-2\pi i x \cdot \xi} \, dx \to 0, \quad \text{as } |\xi| \to \infty.\]
- Use part (a) to justify whether there exists a function \(h \in L^1(\mathbb{R}^d)\) such that
\[f * h = f \quad \text{for all } f \in L^1(\mathbb{R}^d).\]
Here \(f * h\) is the convolution of \(f\) and \(h\) defined by
\[(f * h)(x) = \int_{\mathbb{R}^d} f(x - y) h(y) \, dy.\]
Solution
1. Prove the Riemann–Lebesgue lemma for \(C_c^1\) functions.
Let \(\varphi\in C_c^1(\mathbb R^d)\). For \(\xi\ne0\), choose an index \(j\) such that \[ |\xi_j|\ge \frac{|\xi|}{\sqrt d}. \] Integration by parts in the \(x_j\) variable gives \[ \widehat\varphi(\xi) =\frac{1}{2\pi i\xi_j} \int_{\mathbb R^d}\partial_j\varphi(x)e^{-2\pi i x\cdot\xi}\,dx. \] Therefore \[ |\widehat\varphi(\xi)| \le \frac{\sqrt d}{2\pi|\xi|}\,\|\partial_j\varphi\|_1 \le \frac{\sqrt d}{2\pi|\xi|} \sum_{k=1}^d\|\partial_k\varphi\|_1. \] Hence \[ \widehat\varphi(\xi)\to0 \qquad(|\xi|\to\infty). \]
2. Extend the result to every \(f\in L^1(\mathbb R^d)\).
Fix \(\varepsilon>0\). Since \(C_c^1(\mathbb R^d)\) is dense in \(L^1(\mathbb R^d)\), choose \(\varphi\in C_c^1\) such that \[ \|f-\varphi\|_1<\varepsilon. \] For every \(\xi\), \[ |\widehat f(\xi)-\widehat\varphi(\xi)| \le \|f-\varphi\|_1<\varepsilon. \] By Step 1, \(|\widehat\varphi(\xi)|<\varepsilon\) for all sufficiently large \(|\xi|\). Thus \[ |\widehat f(\xi)|<2\varepsilon \] for all sufficiently large \(|\xi|\). Since \(\varepsilon\) is arbitrary, \[ \boxed{\widehat f(\xi)\to0\quad\text{as }|\xi|\to\infty.} \]
3. Show that no \(L^1\) convolution identity exists.
Suppose \(h\in L^1(\mathbb R^d)\) satisfied \[ f*h=f \qquad\text{for every }f\in L^1(\mathbb R^d). \] Taking Fourier transforms gives \[ \widehat f(\xi)\widehat h(\xi)=\widehat f(\xi) \] for every \(f\) and every \(\xi\).
Take the Gaussian \[ f(x)=e^{-\pi|x|^2}, \] whose Fourier transform is itself and is strictly positive everywhere. Hence \[ \widehat h(\xi)=1 \qquad\text{for every }\xi\in\mathbb R^d. \] But Step 2 applied to \(h\in L^1\) gives \[ \widehat h(\xi)\to0 \qquad(|\xi|\to\infty), \] a contradiction. Therefore no such \(h\in L^1(\mathbb R^d)\) exists.