Exercise 8.19

Question

Exercise 19: 
    Suppose $f$ is a continuous function on $\mathbf R$, $f(x+2\pi)=f(x)$, and $\alpha/\pi$ is irrational. Prove that $$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^Nf(x+n\alpha)=
    \frac{1}{2\pi}\int_{-\pi}^\pi f(t)\,dt$$ for every $x$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

Following the hint, we first show this for trigonometric polynomials. Let $$P(x)=\sum_{m=-M}^Mc_me^{imx}.$$ Then $$\frac{1}{2\pi}\int_{-\pi}^\pi P(t)\,dt=\sum_{m=-M}^Mc_m\frac{1}{2\pi}
    \int_{-\pi}^\pi e^{imt}\,dt=c_0$$ and (noting that $e^{im\alpha}
e 1$ for $m
e 0$ since $\alpha/\pi$ is irrational)
    \begin{align*}
        \frac{1}{N}\sum_{n=1}^NP(x+n\alpha) &= \frac{1}{N}\sum_{n=1}^N\sum_{m=-M}^Mc_me^{im(x+n\alpha)} \
        &= \sum_{m=-M}^Mc_me^{im\alpha}\frac{1}{N}\sum_{n=1}^Ne^{(im\alpha)n} \
        &= c_0+\sum_{\genfrac{}{}{0pt}{}{m=-M}{m
e0}}^Mc_me^{im\alpha}\frac{1}{N}\frac{e^{im\alpha}-e^{im\alpha(N+1)}}{1-e^{im\alpha}} \
        &= c_0+\sum_{\genfrac{}{}{0pt}{}{m=-M}{m
e0}}^Mc_me^{im(x+\alpha)}\frac{1}{N}\frac{1-e^{im\alpha N}}{1-e^{im\alpha}}.
    \end{align*}
    Since for $m
e0$, $$\bigg|\frac{1-e^{im\alpha N}}{1-e^{im\alpha}}\bigg|\le\frac{2}{|1-e^{im\alpha}|}<\infty$$ the limit of the sum on the right-hand side above as $Nightarrow\infty$ is
    0. Hence for every $x$, $$\lim_{Nightarrow\infty}\frac{1}{N}\sum_{n=1}^NP(x+n\alpha)=c_0=\frac{1}{2\pi}\int_{-\pi}^\pi P(t)\,dt.$$

For the general case, let $\varepsilon>0$. By Theorem 8.15 there is a trigonometric polynomial $P$ such that $\big|P(x)-f(x)\big|<\varepsilon/3$ for all real $x$. By the result above, there
    is a positive integer $N_0$ such that for all $N\ge N_0$ $$\Bigg|\frac{1}{N}\sum_{n=1}^NP(x+n\alpha)-\frac{1}{2\pi}\int_{-\pi}^\pi P(t)\,dt\Bigg|\le\frac{\varepsilon}{3}.$$ Hence
    \begin{align*}
        & \Bigg|\frac{1}{N}\sum_{n=1}^Nf(x+n\alpha)-\frac{1}{2\pi}\int_{-\pi}^\pi f(t)\,dt\Bigg| \
        & \qquad\le \frac{1}{N}\sum_{n=1}^N\big|f(x+n\alpha)-P(x+n\alpha)\big| + \Bigg|\frac{1}{N}\sum_{n=1}^NP(x+n\alpha)-\frac{1}{2\pi}\int_{-\pi}^\pi P(t)\,dt\Bigg|\;+ \
        & \qquad\phantom{\le}\;\frac{1}{2\pi}\int_{-\pi}^\pi\big|P(t)-f(t)\big|\,dt \
        & \qquad\le \frac{1}{N}\bigg(\frac{N\varepsilon}{3}\bigg)+\frac{\varepsilon}{3}+\frac{1}{2\pi}\bigg(\frac{2\pi\varepsilon}{3}\bigg) \
        & \qquad= \varepsilon
    \end{align*}

Exercise 8.19

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Exercise 8.19

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