\(\mathrm {Exercise \ \oplus \ Problem } \ 4 \)
\(\mathrm {Exercise \ \oplus \ Problem } \ 4 \) |
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\( \qquad \)你好，这里是我的个人网站数学分析的每周一题栏目(数学分析每周一题，其中数学分析指的是数学中的分析学, 主要包括微积分，实分析，复分析) \(\qquad \ \)——————Alina Lagrange
Show that there does not exist a function \( I \in L^{1}\left(\mathbb{R}^{n}\right) \) such that \(f * I=f\) for all \( f \in L^{1}\left(\mathbb{R}^{n} \right) \)
\(\mathcal{P}roof. \)
$$ \begin{aligned} \hat{f}(\xi) &=\int_{\mathbb{R}^{n}} f(x) e^{-2 \pi i x \cdot \xi} d x \\ &=\int_{\mathbb{R}^{n}} f\left(x-\frac{\xi}{2|\xi|^{2}}\right) e^{-2 \pi i\left(x-\frac{\xi}{2|\xi|^{2}}\right) \cdot \xi} d x \\ &=\int_{\mathbb{R}^{n}} f\left(x-\frac{\xi}{2|\xi|^{2}}\right) e^{-2 \pi i x \cdot \xi} e^{2 \pi i \frac{\xi}{2|\xi|^{2}} \cdot \xi} d x \\ &=\int_{\mathbb{R}^{n}}-f\left(x-\frac{\xi}{2|\xi|^{2}}\right) e^{-2 \pi i x \cdot \xi} d x \\ &=\frac{1}{2}\left(\int_{\mathbb{R}^{n}} f(x) e^{-2 \pi i x \cdot \xi} d x-\right. \left.\int_{\mathbb{R}^{n}} f\left(x-\frac{\xi}{2|\xi|^{2}}\right) e^{-2 \pi i x \cdot \xi} d x\right) \end{aligned} $$ Hence $$ \begin{aligned} |\hat{f}(\xi)| &=\frac{1}{2}\left|\int_{\mathbb{R}^{n}}\left(f(x)-f\left(x-\frac{\xi}{2|\xi|^{2}}\right)\right) e^{-2 \pi i x \cdot \xi} d x\right| \\ & \leq \frac{1}{2} \int_{\mathbb{R}^{n}}\left|\left(f(x)-f\left(x-\frac{\xi}{2|\xi|^{2}}\right)\right) e^{-2 \pi i x \cdot \xi}\right| d x \\ &=\frac{1}{2} \int_{\mathbb{R}^{n}}\left|f(x)-f\left(x-\frac{\xi}{2|\xi|^{2}}\right)\right| d x \end{aligned} $$ By the continuity of \(L^{1}\), and \(\frac{\xi}{2|\xi|^{2}} \rightarrow 0,|\xi| \rightarrow \infty\) If there exists a function \(I \in L^{1}\left(\mathbb{R}^{n}\right)\) such that \(f * I=f\) for all \( f\in L^{1}\left(\mathbb{R}^{n}\right)\). Then \( \hat{f}(\xi) \hat{I}(\xi)=\hat{f}(\xi)\), so we have \(\hat{I}(\xi)=1\) for all \(\xi\). By the proof of \(\hat{f}(\xi) \rightarrow 0 \) as \(|\xi| \rightarrow \infty\). A contradiction.