Exercise 8.21

Question

Exercise 21: 
    Let $$L_n=\frac{1}{2\pi}\int_{-\pi}^\pi\big|D_n(t)\big|\,dt\quad\quad(n=1,2,3,\ldots).$$ Prove that there exists a constant $C>0$ such that $$L_n>C\log n\quad\quad(n=1,2,3,\ldots),$$
    or, more precisely, that the sequence $$\bigg\{L_n-\frac{4}{\pi^2}\log n\bigg\}$$ is bounded.

analambanomenos · Accepted Answer

Since $D_n$ is an even function, and making a substitution, we can simplify the integral to
    \begin{align*}
        L_n(x) &= \frac{1}{2\pi}\int_{-\pi}^\pi\big|D_n(x)\big|\,dx \
        &= \frac{1}{\pi}\int_0^\pi\frac{\big|\sin\big(n+\frac{1}{2}\big)x\big|}{\sin(x/2)}\,dx \
        &= \frac{2}{\pi}\int_0^{\pi/2}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx
    \end{align*}
    Letting $A=\pi/(2N+1)$, we can break up the interval of integration into the subintervals $$I_1=[0,A]\quad I_2=[A,2A]\quad\cdots\quad I_n=\big[(n-1)A,nA\big]\quad\big[nA,\pi/2\big]$$ each
    having length $A$ except the last, which has length $A/2$ and which we can ignore for purposes of the estimate. In the interior of the interval $I_m$, $\sin(2n+1)x$ is positive for $m$ odd,
    and negative for $m$ even. Also, $\sin x$, which is positive and monotonically increasing over $I_m$, has the maximum value $\sin(mA)<mA$. Hence, for $m$ odd, we have
    \begin{align*}
        \frac{2}{\pi}\int_{(m-1)A}^{mA}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx &\ge \frac{2}{\pi}\cdot\frac{1}{mA}\int_{(m-1)A}^{mA}\sin(2n+1)x\,dx \
        &= \frac{2}{\pi}\cdot\frac{2n+1}{m\pi}\cdot\frac{1}{2n+1}\int_{(m-1)\pi}^{m\pi}\sin x\,dx \
        &= \frac{2}{\pi^2}\cdot\frac{1}{m}\Big(\cos\big((m-1)\pi\big)-\cos(m\pi)\Big) \
        &= \frac{4}{\pi^2}\cdot\frac{1}{m}
    \end{align*}
    and for $m$ even, we have
    \begin{align*}
        \frac{2}{\pi}\int_{(m-1)A}^{mA}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx &\ge \frac{2}{\pi}\cdot\frac{1}{mA}\int_{(m-1)A}^{mA}-\sin(2n+1)x\,dx \
        &= \frac{2}{\pi}\cdot\frac{2n+1}{m\pi}\cdot\frac{1}{2n+1}\int_{(m-1)\pi}^{m\pi}-\sin x\,dx \
        &= \frac{2}{\pi^2}\cdot\frac{1}{m}\Big(\cos(m\pi)-\cos\big((m-1)\pi\big)\Big) \
        &= \frac{4}{\pi^2}\cdot\frac{1}{m}
    \end{align*}
    Recall that in Exercise 9 it was shown that $1+(1/2)+\cdots+(1/n)>\log n$. Hence, adding the inequalities above for $m=1,2,\ldots,n$, we get $$L_n\ge\frac{4}{\pi^2}\bigg(1+\frac{1}{2}+\cdots+
    \frac{1}{n}\bigg)>\frac{4}{\pi^2}\log n$$ which shows the first part of the Exercise, with $C=4/\pi^2$. (To solve the second part, you would have to get an upper estimate for $L_n$).

Exercise 8.21

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Exercise 8.21

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