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We denote by $L^1(R^n)$ the space of Lebesgue integrable functions on $R^n$. For $f\in L^1(R^n)$, the Fourier transformation $\widehat{f}$ of $f$ is defined by
$$\widehat{f}(\xi)=\int f(x)e^{i\xi \cdot x}dx, \quad \xi \in R^n.$$
Fact 1. $\widehat{f}(\xi)$ is uniformly continuous on $R^n.$
Fact 2. If $h(x)=f(x) \ast g(x):=\int f(x-y)g(y)dy,$ then $\widehat{h}(\xi)=\widehat{f}(\xi)\cdot \widehat{g}(\xi).$
Fact 3. Let $f\in L^1(R^n)$ and $xf(x)\in L^1(R^n)$. Then $\widehat(f)$ is differentiable and $\frac{d}{d\xi}\widehat{f}(\xi)=\widehat{(-ixf)}(\xi).$
Fact 4. (Riemann-Lebesgue Lemma) For $f\in L^1(R^n)$, $\lim\limits_{|\xi|\to 0}\widehat{f}(\xi)=0.$
Important Example: Let $f(x)=|x|^{-s}, x\in R^n$. Then $\widehat{f}(\xi)=c(s,n)|x|^{s-n}.$ See "Lectures in Harmonic analysis" by Thomas H. Wolff.
We denote by $M(R^n)$ the space of all finite Borel measures on $R^n.$ $M(R^n)$ is identified with the dual space of $C_0(R^n)$--the (sup-normed) space of all continuous fnctions on $R^n$ which vanish at infity--by means of the coupling
$$<\mu, f>=\int fd\mu.\quad f\in C_0(R^n), \mu\in M(R^n).$$
The convolution of a measure $\mu\in M(R^n)$ and a function $\varphi\in C_0(R^n)$ is defined by
$$(\varphi \ast\mu)(x)=\int \varphi(x-y)d\mu(y).$$
Clearly, $\varphi\ast\mu\in C_0(R^n)$ and $\|\varphi\ast\mu\|\le \|\mu\|\|\varphi\|.$
The Fourier transformation of a measure $\mu\in M(R^n)$ is defined by
$$\widehat{\mu}(\xi)=\int e^{i\xi\cdot x}d\mu(x).$$
Remark: If $\mu \ll \mathcal{L}^n,$ say $d\mu=f(x)dx,$ then $\widehat{\mu}(\xi)=\widehat{f}(\xi).$
Fact 5 (Parseval‘s Theorem) Let $\mu \in M(R^n)$ and let $f$ be a continuous function in $L^1(R^n)$ such that $\widehat{f}\in L^1(R^n)$. Then
$$\int f(x)d\mu(x)=(2\pi)^n\int \widehat{f}(\xi)\widehat{\mu}(-\xi)dx=(2\pi)^n\int \widehat{f}(\xi)\overline{\widehat{\mu}(\xi)}dx.$$
Some useful facts on Fourier transformation
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原文地址:http://www.cnblogs.com/jinjun/p/5041741.html