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