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Question

허락하다 $\varphi$ 다음과 같은 테스트 기능 $Supp(\varphi)\subseteq [-M, M]$, $a\in \mathbb R$. 우리는 분배 원칙 값을 정의합니다.$\frac{1}{x-a}$ :

$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right),\varphi\right\rangle: =\lim_{\varepsilon\to 0^+}\int_{|x - a|\geq \varepsilon}\frac{\varphi(x)}{x-a} \, \mathrm{d}x$

어떻게 표시합니까?

$$\lim_{a\to 0} \frac{1}{2a}\left(\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right)\right) = \operatorname{F.\!p.}\left(\frac{1}{x^2}\right)$$

나는 증명했다 : $$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right), \varphi \right\rangle = \lim_{\varepsilon\to 0^+}\int_{ \varepsilon}^M\frac{2\varphi(x)-\varphi(2a - x) - \varphi(x-2a)}{x-a} \, \mathrm{d}x $$

제 추론이 맞습니까? 도움을 주셔서 감사합니다.

Answer 1

알고 있다면 $\left( \operatorname{pv}\frac{1}{x} \right)' = -\operatorname{fp}\frac{1}{x^2}$: $$ \left< \frac{1}{2a} \left( \operatorname{pv}\frac{1}{x-a} - \operatorname{pv}\frac{1}{x+a} \right), \varphi(x) \right> = \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x-a}, \varphi(x) \right> - \left< \operatorname{pv}\frac{1}{x+a}, \varphi(x) \right> \right) \\ = \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x}, \varphi(x+a) \right> - \left< \operatorname{pv}\frac{1}{x}, \varphi(x-a) \right> \right) = \left< \operatorname{pv}\frac{1}{x}, \frac{\varphi(x+a) - \varphi(x-a)}{2a} \right> \\ \to \left< \operatorname{pv}\frac{1}{x}, \varphi'(x) \right> = - \left< \left( \operatorname{pv}\frac{1}{x} \right)', \varphi(x) \right> = \left< \operatorname{fp}\frac{1}{x^2}, \varphi(x) \right> $$