【第八章 习题21】
设
Ln=12π∫−ππ∣Dn(t)∣dt (n=1,2,3,...)L_{n} = \frac{1}{2\pi}\int_{- \pi}^{\pi}{\left| D_{n}(t) \right|dt}\ \ \ \ \ \ (n = 1,2,3,\ldots)Ln=2π1∫−ππ∣Dn(t)∣dt (n=1,2,3,...)
试证，存在着常数C>0C > 0C>0，使得
Ln>Clogn (n=1,2,3,...)L_{n} > C\log n\ \ \ \ \ (n = 1,2,3,\ldots)Ln>Clogn (n=1,2,3,...)
或者更精确些，就是序列
{Ln−4π2logn}\left\{ L_{n} - \frac{4}{\pi^{2}}\log n \right\}{Ln−π24logn}
有界。
【证明】
根据迪利克雷核的定义，可知
Ln=1π∫02π2n+1sin(n+12)tsin12tdt+∑k=1n−11π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsin12tdt+1π∫2nπ2n+1πsin(n+12)tsin12tdtL_{n} = \frac{1}{\pi}\int_{0}^{\frac{2\pi}{2n + 1}}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt} + \sum_{k = 1}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}} + \frac{1}{\pi}\int_{\frac{2n\pi}{2n + 1}}^{\pi}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}Ln=π1∫02n+12πsin21tsin(n+21)tdt+k=1∑n−1π1∫2n+12πk2n+12π(k+1)sin21t(−1)ksin(n+21)tdt+π1∫2n+12nππsin21tsin(n+21)tdt
由于
limn→∞1π∫02π2n+1sin(n+12)tsin12tdt\lim_{n \rightarrow \infty}{\frac{1}{\pi}\int_{0}^{\frac{2\pi}{2n + 1}}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}}n→∞limπ1∫02n+12πsin21tsin(n+21)tdt
=u=(n+12)tlimn→∞1π∫0πsinu(n+12)sinu2n+1du\overset{u = \left( n + \frac{1}{2} \right)t}{=}\lim_{n \rightarrow \infty}{\frac{1}{\pi}\int_{0}^{\pi}{\frac{\sin u}{\left( n + \frac{1}{2} \right)\sin\frac{u}{2n + 1}}du}}=u=(n+21)tn→∞limπ1∫0π(n+21)sin2n+1usinudu
=limn→∞12π∫0πsinu(2n+1)sinu2n+1du= \lim_{n \rightarrow \infty}{\frac{1}{2\pi}\int_{0}^{\pi}{\frac{\sin u}{(2n + 1)\sin\frac{u}{2n + 1}}du}}=n→∞lim2π1∫0π(2n+1)sin2n+1usinudu
=12π∫0πsinuudu=I= \frac{1}{2\pi}\int_{0}^{\pi}{\frac{\sin u}{u}du} = I=2π1∫0πusinudu=I
其中sinu(2n+1)sinu2n+1\frac{\sin u}{(2n + 1)\sin\frac{u}{2n + 1}}(2n+1)sin2n+1usinu一致收敛到sinuu\frac{\sin u}{u}usinu。
limn→∞1π∫2nπ2n+1πsin(n+12)tsin12tdt=u=(n+12)tlimn→∞1π∫nπ(n+12)πsinu(n+12)sinu2n+1du≤limn→∞1π∫nπ(n+12)πsinu(n+12)sinnπ2n+1du=0{\lim_{n \rightarrow \infty}{\frac{1}{\pi}\int_{\frac{2n\pi}{2n + 1}}^{\pi}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}} }{\overset{u = \left( n + \frac{1}{2} \right)t}{=}\lim_{n \rightarrow \infty}{\frac{1}{\pi}\int_{n\pi}^{\left( n + \frac{1}{2} \right)\pi}{\frac{\sin u}{\left( n + \frac{1}{2} \right)\sin\frac{u}{2n + 1}}du}} \leq \lim_{n \rightarrow \infty}{\frac{1}{\pi}\int_{n\pi}^{\left( n + \frac{1}{2} \right)\pi}{\frac{\sin u}{\left( n + \frac{1}{2} \right)\sin\frac{n\pi}{2n + 1}}du} = 0}}n→∞limπ1∫2n+12nππsin21tsin(n+21)tdt=u=(n+21)tn→∞limπ1∫nπ(n+21)π(n+21)sin2n+1usinudu≤n→∞limπ1∫nπ(n+21)π(n+21)sin2n+1nπsinudu=0
所以
Ln=I+ε1+∑k=1n−11π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsin12tdtL_{n} = I + \varepsilon_{1} + \sum_{k = 1}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}}Ln=I+ε1+k=1∑n−1π1∫2n+12πk2n+12π(k+1)sin21t(−1)ksin(n+21)tdt
类似地
Ln=1π∫02π2n+1sin(n+12)tsin12tdt+1π∫2π2n+14π2n+1−sin(n+12)tsin12tdt+∑k=2n−11π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsin12tdt+1π∫2nπ2n+1πsin(n+12)tsin12tdt=L_{n} = \frac{1}{\pi}\int_{0}^{\frac{2\pi}{2n + 1}}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt} + \frac{1}{\pi}\int_{\frac{2\pi}{2n + 1}}^{\frac{4\pi}{2n + 1}}{\frac{- \sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt} + \sum_{k = 2}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}} + \frac{1}{\pi}\int_{\frac{2n\pi}{2n + 1}}^{\pi}{\frac{\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt} =Ln=π1∫02n+12πsin21tsin(n+21)tdt+π1∫2n+12π2n+14πsin21t−sin(n+21)tdt+k=2∑n−1π1∫2n+12πk2n+12π(k+1)sin21t(−1)ksin(n+21)tdt+π1∫2n+12nππsin21tsin(n+21)tdt=
Ln=2I+ε2+∑k=2n−11π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsin12tdtL_{n} = 2I + \varepsilon_{2} + \sum_{k = 2}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}}Ln=2I+ε2+k=2∑n−1π1∫2n+12πk2n+12π(k+1)sin21t(−1)ksin(n+21)tdt
由于
1π∫2πk2n+12π(k+1)2n+12sinπ(k+1)2n+1dt=1π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsinπ(k+1)2n+1dt\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\sin\frac{\pi(k + 1)}{2n + 1}}dt} = \frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin\frac{\pi(k + 1)}{2n + 1}}dt}π1∫2n+12πk2n+12π(k+1)sin2n+1π(k+1)2dt=π1∫2n+12πk2n+12π(k+1)sin2n+1π(k+1)(−1)ksin(n+21)tdt
<1π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsin12tdt< \frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin{\frac{1}{2}t}}dt}<π1∫2n+12πk2n+12π(k+1)sin21t(−1)ksin(n+21)tdt
<1π∫2πk2n+12π(k+1)2n+1(−1)ksin(n+12)tsinπk2n+1dt=1π∫2πk2n+12π(k+1)2n+12sinπk2n+1dt< \frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{( - 1)^{k}\sin{\left( n + \frac{1}{2} \right)t}}{\sin\frac{\pi k}{2n + 1}}dt} = \frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\sin\frac{\pi k}{2n + 1}}dt}<π1∫2n+12πk2n+12π(k+1)sin2n+1πk(−1)ksin(n+21)tdt=π1∫2n+12πk2n+12π(k+1)sin2n+1πk2dt
又因为
∣1sinx−1x∣<Kx\left| \frac{1}{\sin x} - \frac{1}{x} \right| < Kx sinx1−x1 <Kx
所以
∑k=1n−11π∫2πk2n+12π(k+1)2n+12sinπ(k+1)2n+1dt=En+∑k=1n−11π∫2πk2n+12π(k+1)2n+12π(k+1)2n+1dt=∑k=1n−14π21k+1\sum_{k = 1}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\sin\frac{\pi(k + 1)}{2n + 1}}dt}} = E_{n} + \sum_{k = 1}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\frac{\pi(k + 1)}{2n + 1}}dt}} = \sum_{k = 1}^{n - 1}{\frac{4}{\pi^{2}}\frac{1}{k + 1}}k=1∑n−1π1∫2n+12πk2n+12π(k+1)sin2n+1π(k+1)2dt=En+k=1∑n−1π1∫2n+12πk2n+12π(k+1)2n+1π(k+1)2dt=k=1∑n−1π24k+11
∑k=2n−11π∫2πk2n+12π(k+1)2n+12sinπk2n+1dt=Fn+∑k=2n−11π∫2πk2n+12π(k+1)2n+12πk2n+1dt=∑k=2n−14π21k\sum_{k = 2}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\sin\frac{\pi k}{2n + 1}}dt}} = F_{n} + \sum_{k = 2}^{n - 1}{\frac{1}{\pi}\int_{\frac{2\pi k}{2n + 1}}^{\frac{2\pi(k + 1)}{2n + 1}}{\frac{2}{\frac{\pi k}{2n + 1}}dt}} = \sum_{k = 2}^{n - 1}{\frac{4}{\pi^{2}}\frac{1}{k}}k=2∑n−1π1∫2n+12πk2n+12π(k+1)sin2n+1πk2dt=Fn+k=2∑n−1π1∫2n+12πk2n+12π(k+1)2n+1πk2dt=k=2∑n−1π24k1
其中
En<1πK22n+1∑k=1n−1π(k+1)2n+1=2K(2n+1)212n(n+1)−1≈K4E_{n} < \frac{1}{\pi}K\frac{2}{2n + 1}\sum_{k = 1}^{n - 1}\frac{\pi(k + 1)}{2n + 1} = \frac{2K}{(2n + 1)^{2}}\left\lbrack \frac{1}{2}n(n + 1) - 1 \right\rbrack \approx \frac{K}{4}En<π1K2n+12k=1∑n−12n+1π(k+1)=(2n+1)22K21n(n+1)−1≈4K
Fn<1πK22n+1∑k=2n−1πk2n+1=2K(2n+1)212n(n−1)−1≈K4F_{n} < \frac{1}{\pi}K\frac{2}{2n + 1}\sum_{k = 2}^{n - 1}\frac{\pi k}{2n + 1} = \frac{2K}{(2n + 1)^{2}}\left\lbrack \frac{1}{2}n(n - 1) - 1 \right\rbrack \approx \frac{K}{4}Fn<π1K2n+12k=2∑n−12n+1πk=(2n+1)22K21n(n−1)−1≈4K
所以
I+ε1+K4+δ1+∑k=1n−14π21k+1<Ln<2I+ε2+K4+δ2+∑k=2n−14π21kI + \varepsilon_{1} + \frac{K}{4} + \delta_{1} + \sum_{k = 1}^{n - 1}{\frac{4}{\pi^{2}}\frac{1}{k + 1}} < L_{n} < 2I + \varepsilon_{2} + \frac{K}{4} + \delta_{2} + \sum_{k = 2}^{n - 1}{\frac{4}{\pi^{2}}\frac{1}{k}}I+ε1+4K+δ1+k=1∑n−1π24k+11<Ln<2I+ε2+4K+δ2+k=2∑n−1π24k1
从而
I+ε1+K4+δ1−4π2<Ln−∑k=1n4π21k<2I+ε2+K4+δ2−4π2−4π21nI + \varepsilon_{1} + \frac{K}{4} + \delta_{1} - \frac{4}{\pi^{2}} < L_{n} - \sum_{k = 1}^{n}{\frac{4}{\pi^{2}}\frac{1}{k}} < 2I + \varepsilon_{2} + \frac{K}{4} + \delta_{2} - \frac{4}{\pi^{2}} - \frac{4}{\pi^{2}}\frac{1}{n}I+ε1+4K+δ1−π24<Ln−k=1∑nπ24k1<2I+ε2+4K+δ2−π24−π24n1
而
0<∑k=1n1k−logn<10 < \sum_{k = 1}^{n}\frac{1}{k} - \log n < 10<k=1∑nk1−logn<1
所以
Ln−∑k=1n4π2lognL_{n} - \sum_{k = 1}^{n}{\frac{4}{\pi^{2}}\log n}Ln−k=1∑nπ24logn
是有上下界的。