实分析作业-第四周

1. 若$fg$均为周期$2\pi$的可积函数，定义$f\ast g(x)=\frac{1}{2\pi }{\int }_0^{2\pi }f(x-y)g(y)dy$，证明：
    (a). $f\ast g(x)$连续
    (b).$\widehat{f\ast g}(n)=\hat{f}(n)\hat{g}(n)$
   $$\begin{array}{rl}(a).& |f\ast g(x+\delta )-f\ast g(x)|\\ =& \frac{1}{2\pi }|{\int }_0^{2\pi }g(y)[f(x+\delta -y)-f(x-y)]dy|\\ \le & \sqrt{\frac{1}{(2\pi {)}^2}{\int }_0^{2\pi }|g(y){|}^{2}dy{\int }_0^{2\pi }|f(x+\delta -y)-f(x-y){|}^{2}dy}\\ \le & |\underset{[0,2\pi ]}{\sup }g|\sqrt{\frac{1}{2\pi }{\int }_0^{2\pi }|f(x+\delta -y)-f(x-y){|}^{2}dy}\\ & 又f可积，故f至多有有限个间断点\\ & \therefore \forall x-y,\exists 足够小的\delta ,\mathbf{s}.\mathbf{t}.f(x+\delta -y)-f(x-y)\to 0\\ & \therefore \sup [f(x+\delta -y)-f(x-y)]\to 0(\delta \to 0)\\ & \therefore {\int }_0^{2\pi }|f(x+\delta -y)-f(x-y){|}^{2}dy\to 0(\delta \to 0)\\ & \therefore |f\ast g(x+\delta )-f\ast g(x)|\to 0(\delta \to 0)\\ & 即f\ast g(x)连续\\ (b).& \widehat{f\ast g}(n)\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }f\ast g(x)e^{-inx}dx\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }{\int }_0^{2\pi }f(x-y)g(y)e^{-inx}dydx\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }g(y)e^{-iny}[\frac{1}{2\pi }{\int }_0^{2\pi }f(x-y)e^{-in(x-y)}dx]dy\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }g(y)e^{-iny}[\frac{1}{2\pi }{\int }_0^{2\pi }f(x-y)e^{-in(x-y)}d(x-y)]dy\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }g(y)e^{-iny}[\frac{1}{2\pi }{\int }_{0-y}^{2\pi -y}f(x)e^{-inx}dx]dy\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }g(y)e^{-iny}[\frac{1}{2\pi }{\int }_0^{2\pi }f(x)e^{-inx}dx]dy\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }g(y)e^{-iny}dy[\frac{1}{2\pi }{\int }_0^{2\pi }f(x)e^{-inx}dx]\\ =& \hat{f}(n)\hat{g}(n)\end{array}$$

 ( a). 说明Good Kernel的性质
 ( b).证明Theorem4.1

 ( c). 证明fejer kernel 是good kernel(lemma 5.1)
$$\begin{array}{rl}(a).& 1.\forall n\ge 1,\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(x)dx=1\\ & 2.\exists M>0,使得对\forall n\ge 1,{\int }_{-\pi }^{\pi }|K_n(x)|dx\le M\\ & 3.对所有\delta >0,有{\int }_{\delta \le |x|\le \pi }|K_n(x)|dx\to 0(n\to \infty )\\ (b).& |f\ast K_n(x)-f(x)|\\ \le & \frac{1}{2\pi }{\int }_{-\pi }^{\pi }|[f(x-y)-f(x)]K_n(y)|dy\\ =& \frac{1}{2\pi }{\int }_{|y|<\delta }|[f(x-y)-f(x)]K_n(y)|dy+\frac{1}{2\pi }{\int }_{\delta \le |y|\le \pi }|[f(x-y)-f(x)]K_n(y)|dy\\ \le & \frac{\sup [f(x-y)-f(x)]2\delta }{2\pi }{\int }_{|y|<\delta }|K_n(y)|dy+\frac{\sup [f(x-y)-f(x)]2[\pi -\delta ]}{2\pi }{\int }_{\delta \le |y|\le \pi }|K_n(y)|dy\\ \le & \frac{\sup [f(x-y)-f(x)]2\delta }{2\pi }M+\frac{\sup [f(x-y)-f(x)]2[\pi -\delta ]}{2\pi }{\int }_{\delta \le |y|\le \pi }|K_n(y)|dy\\ \to & 0(\delta ,n\to 0)\end{array}$$

3. 习题2.7
   
   $$\begin{array}{rlrl}(a).& 当N=M+1时，& & \\ & \sum _{n=M}^N{a}_n{b}_n& =& a_M{b}_M+a_{M+1}{b}_{M+1}\\ & & =& a_N(B_N-B_M)+a_M(B_M-B_{M-1})\\ & & =& a_N{B}_N-a_M{B}_{M-1}-(a_N-a_M)B_M\\ & & =& a_N{B}_N-a_M{B}_{M-1}-\sum _{n=M}^{N-1}(a_{n+1}-a_n)B_n\\ & 当N>M+1时，& & \\ & \sum _{n=M}^N{a}_n{b}_n& =& a_N{b}_N+\sum _{n=M}^{N-1}{a}_n{b}_n\\ & & =& a_N(B_N-B_{N-1})+a_{N-1}{B}_{N-1}-a_M{B}_M-\sum _{n=M}^{N-2}(a_{n+1}-a_n)B_n\\ & & =& a_N{B}_N+a_{N-1}{B}_{N-1}-a_M{B}_M-\sum _{n=M}^{N-1}(a_{n+1}-a_n)B_n\\ & 证毕& & \\ (b).& 令S_N=\sum _{n=0}^N{a}_n{b}_n,B_n\le B& & \\ & 则\underset{n\to \infty }{\lim }{S}_N& =& a_N{B}_N+a_{N-1}{B}_{N-1}-\sum _{n=0}^{N-1}(a_{n+1}-a_n)B_n\\ & & \le & a_{N}B+a_{N-1}B-\sum _{n=0}^{N-1}(a_{n+1}-a_n)B\\ & & =& B[\sum _{n=0}^N{a}_n-\sum _{n=1}^{N-1}{a}_n+a_{N-1}]\\ & & =& B[a_{n-1}-a_0]<\infty \\ & 故级数收敛& & \end{array}$$
4. 利用上一题的结论，证明${\sum }_{n\ne 0}\frac{e^{inx}}{n}$对每一个x都是收敛的
   $$\begin{array}{rl}& 将级数分为\sum _{n>0}\frac{e^{inx}}{n}与\sum _{n>0}\frac{1}{n}{e}^{-inx}\\ \sum _{n>0}\frac{e^{inx}}{n}=& \sum _{n>0}\frac{\cos nx+i\sin nx}{n}\\ =& \sum _{n>0}\frac{\cos nx}{n}+i\sum _{n>0}\frac{\sin nx}{n}\\ & 又\sum _{k=1}^N\cos kx,\sum _{k=1}^N\sin kx均在x\ne \frac{n\pi }{2}时有界,\frac{1}{n}单减到0\\ & 故\sum _{n>0}\frac{\cos nx}{n}，\sum _{n>0}\frac{\sin nx}{n}均收敛\\ & \therefore \sum _{n>0}\frac{e^{inx}}{n}收敛\\ \sum _{n>0}\frac{1}{n}{e}^{-inx}=& \sum _{n>0}\frac{\cos (-nx)+i\sin (-nx)}{n}同样收敛\\ & 因此\sum _{n\ne 0}\frac{e^{inx}}{n}收敛\end{array}$$