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Let $0 < \alpha_1 < \alpha_2 < \cdots < \alpha_k < 1$, and $a_1,\ldots , a_k \in \mathbb{R}$. Suppose that $$\lim_{n\to + \infty} \sum_{p = 1}^k a_p \sin (n \pi \alpha_p)=0$$

Show that $a_1=a_2=\cdots=a_k=0$.

I tried to use the same method as Cantor Lemma which you can find a related problem on MSE : Convergence of series of functions: $f_n(x)=u_n\sin(nx)$

But I failed, it seems to be more difficult.

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  • $\begingroup$ I haven't actually confirmed it, but can't this be done "simply" using induction on $k$? $\endgroup$ – Mark Hurd May 20 '14 at 1:24
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    $\begingroup$ Strictly, that would be a contradiction. Your initial assumption is that all $a_i$ are positive, so you cannot prove that they are all zero. $\endgroup$ – user3294068 May 27 '14 at 17:25
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    $\begingroup$ @user3294068 the coefficients are latin. the frequencies are greek. $\endgroup$ – Robert Wolfe May 27 '14 at 17:31
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    $\begingroup$ You should look up literature on almost-periodic functions. I was previously unaware of their theory, but I learned some interesting things. Unlike periodic functions, they actually form a vector space over $\Bbb R$. In addition, they are the the uniform closure of the periodic functions under the supremum norm. This in particular says that your function can be uniformly approximated by a periodic function. Thus if its limit is zero, it would be identically zero. Since $\sin$ functions of different frequencies are linearly independent, the $a_k$ are 0. I couldn't come up with a simpler answer. $\endgroup$ – Robert Wolfe May 27 '14 at 18:10
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    $\begingroup$ The specifics of your problem hint that there might be a more elementary solution though. What is the source of this problem? $\endgroup$ – Robert Wolfe May 27 '14 at 18:18
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We will use the next lemma:

Lemma. For $\alpha,\beta$ from $(0,1)$ and a positive integer $n$, let $$ I_n(\alpha,\beta)=\frac{2}{n}\sum_{k=1}^n\sin(\alpha \pi k)\sin(\beta \pi k).$$ Then $$\lim_{n\to\infty}I_n(\alpha,\beta)=\left\{\matrix{0&\hbox{if}&\alpha\ne \beta,\cr 1&\hbox{if}&\alpha=\beta.} \right. $$

Proof. Note that, for $\theta\notin 2\pi\mathbb{Z} $, we have $$ \sum_{k=1}^n\cos(k\theta)=-\frac{1}{2}+\frac{1}{2}\sum_{k=-n}^ne^{ik\theta} =-\frac{1}{2}+\frac{1}{2}\frac{e^{i(n+1)\theta}-e^{-in\theta}}{e^{i\theta}-1} =-\frac{1}{2}+\frac{\sin((n+1/2)\theta)}{2\sin(\theta/2)} $$ So $$\forall\,\theta\in\mathbb{R}\setminus2\pi\mathbb{Z}, \quad\lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n\cos(k\theta)=0\tag{1} $$ Now, if $\alpha\ne\beta$ $$ I_n(\alpha,\beta)=\frac{1}{n}\sum_{k=1}^n\cos(k\pi(\alpha-\beta)) -\frac{1}{n}\sum_{k=1}^n\cos(k\pi(\alpha+\beta)) $$ and by $(1)$, $\lim_{n\to\infty}I_n(\alpha,\beta)=0$ in this case. Similarly, if $\alpha=\beta$, we have $$ I_n(\alpha,\alpha)=\frac{1}{n}\sum_{k=1}^n(1-\cos(2k\pi\alpha)) =1-\frac{1}{n}\sum_{k=1}^n\cos(2k\pi\alpha) $$ so, $\lim_{n\to\infty}I_n(\alpha,\alpha)=1$.$\qquad\square$

Now, let us come the the proposed question. Let $$ x_n=\sum_{p=1}^ka_p\sin(n\pi\alpha_p) $$ by assumption $\lim\limits_{n\to\infty}x_n=0$, So, for $q\in\{1,2,\ldots,k\}$ we have $\lim\limits_{n\to\infty}x_n\sin(n\pi\alpha_q)=0$. Using Stolz–Cesàro theorem we conclude that $$ \lim_{n\to\infty}\frac{2}{n}\sum_{m=1}^nx_m\sin(m\pi\alpha_q)=0 $$ This is equivalent to $$ \lim_{n\to\infty}\sum_{p=1}^ka_pI_n(\alpha_p,\alpha_q)=0 $$ or $a_q=0$ according to the Lemma, and the desired conclusion follows.$\qquad\square$

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  • $\begingroup$ Wah! So clever. (+1) $\endgroup$ – user146010 May 27 '14 at 18:55
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There is also a one-line solution through diophantine approximation.

The sequence $\left\{\left(\left\{\frac{n\alpha_1}{2}\right\},\ldots,\left\{\frac{n\alpha_k}{2}\right\}\right)\right\}_{n\in\mathbb{N}}$ is ergodic in $[0,1]^k$, hence is some $a_p$ differs from zero, $\{x_n=\sum_{p=1}^{k}a_p \sin(\pi n \alpha_p)\}_{n\in\mathbb{N}}$ has a non-Cauchy subsequence and cannot converge.

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This approach picks up on the comment I made. After that reading, I came up with an alternate way to prove this.

We have in general that the sum of two periodic functions is not in general periodic. However, when the periods of the summands are commensurable, their sum will also be periodic. Take a look at this question. My solution exploits this fact and the idea that changing the frequency of a wave by a tiny amount will still approximate a wave fairly well, as long as we don't look too far ahead. As an example, $\cos 2.0001x$ approximates $\cos 2x$ really well for a long time. In fact, you need to look at $x$ greater than $1000$ to see the former finally miss the first decimal place in the approximation.

These two facts imply that we can create a sequence of periodic functions $f_n$ such that $f_n$ converges to $f$ compactly. Look at this article. We can also choose the $f_n$ to only improve their approximation on each interval of the form $[-n, n]$. See the animation here.

Now, we know that $f$ attains a non-zero value somewhere, say at $x_0$ (since $\sin$ functions of different frequency are linearly independent). Call this value $a$. Now suppose we had that $f$ approached $0$ as $x\rightarrow\infty$. This means that there is an $M\in\Bbb N$ such that $|f(x)|<|a|/2$ for all $x>M$. Now since the $f_n$ approximate $f$ compactly there is eventually an $f_N$ such that $\sup|f(x)-f_k(x)|<|a|/2$ for all $k\geq N$ and $x<M+200$ (the 200 is arbitrary; just need something bigger than 0). But $f_N$ is periodic. This means its amplitude is less than $|a|/2 $ everywhere. This contradicts the fact that our earlier $f_n$ approximated $f$ well around $x_0$.

Basically, our early terms of the sequence $f_n$ had a 'large' amplitude. If $f$ approached zero, our later terms would have to decrease in amplitude, and this throws off the fact we chose the $f_n$ to only improve their approximation as we went out further.

EDIT

Knowing what my $f_m$ (re-indexed $n$ to $m$) are would help. Each $f_m$ is of the form $$\sum_{j=1}^k a_{j}\sin(\alpha_{j_m}\pi n)$$ where the $\alpha_{j_m}$ are chosen so that $\alpha_{j_m}$ is 'sufficiently close' to $\alpha_j$ and we have that $\alpha_{1_m}, \dots, \alpha_{k_m}$ are commensurable.

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