【转载请注明出处】http://www.cnblogs.com/mashiqi

2017/02/20

在$\mathcal{D}(0,1)$上取定$\varphi_0 \in \mathcal{D}(0,1)$满足$\int_0^1 \varphi_0(x) \mathrm{d}x = 1$。令$$\phi_{\varphi}(x) \overset{\Delta}{=} \varphi(x) - \int_0^1 \varphi(t) \mathrm{d}t \cdot \varphi_0(x), ~\text{where}~ \varphi \in \mathcal{D}(0,1).$$则我们有$$\forall \varphi \in \mathcal{D}(0,1), ~\text{we have}~ \phi_{\varphi}(x) \in \mathcal{D}(0,1) ~\text{and}~ \Phi_{\varphi}(x) \overset{\Delta}{=} \int_0^x \phi_{\varphi}(t) \in \mathcal{D}(0,1).$$

我们应该注意到$$\forall \phi \in \mathcal{D}(0,1), \phi \in \mathcal{D}(0,1) \Rightarrow \phi' \in \mathcal{D}(0,1)$$但上述关系一般不能反过来，即从$\phi \in \mathcal{D}(0,1)$一般是得不到$\int_0^x \phi(t)\mathrm{d}t \in \mathcal{D}(0,1)$的。但是上面构造出来的$\phi_{\varphi}$既满足$\phi_{\varphi}(x) \in \mathcal{D}(0,1)$又满足$\int_0^x \phi_{\varphi}(t) \in \mathcal{D}(0,1).$这个结论还挺有趣。