Is there a sense in which this limit is zero?

This question arose from this answer of mine on DSP.SE. Basically, I need to check if this is true:

$$\lim_{M\to \infty} j\frac{\cos[\omega(M+1/2)]}{2\sin(\omega/2)}\stackrel{?}{=}0$$

where $j$ is the imaginary unit (I don't think it's relevant though, we could just drop it), $M\in\mathbb{Z}$ and $\omega\in\mathbb{R}$.

Note that we are talking about the sequence of functions here, not typical limits.

This doubt came to me because if the cosine in the numerator was replaced by a sine, then the sequence would converge to $j\pi\delta(\omega)$, where $\delta$ denotes the Dirac delta function. This made me think that maybe there was something similar that could be found out in this case (with the cosine).

I don't know whether this is right or not, but I feel that that limit has to be the zero distribution (if you're interested in the reason, you can check out the link above). Does anyone have any idea?

• Does $M\to\infty$ as an integer? as a real? What is $\omega$? can it be any real? – robjohn Jan 23 '18 at 1:21
• @robjohn Sorry for not pointing those things out, I've edited the question. – Tendero Jan 23 '18 at 1:23
• Judging from comments, I think you mean something involving distributions. This is context that really really needs to be expressed! In mathematical settings, the default is usually to assume that you are speaking about the limits of functions, like what you learn in introductory calculus. – Hurkyl Jan 23 '18 at 1:25
• @robjohn Based on the comments to my answer, I think that the question is trying to ask: Consider the sequence $f_n= j\frac{\cos[\omega(n+1/2)]}{2\sin(\omega/2)}$. Do they converge, as distributions to the zero distribution? Which is something completely different than I understood :) – N. S. Jan 23 '18 at 1:34

Let us interpret functions in question as the following distributions defined in principal-value sense

$$\forall \varphi \in C_c^{\infty}(\mathbb{R}) \ : \quad T_M(\varphi) = \mathrm{P.V.}\int_{-\infty}^{\infty} \frac{\cos((M+\frac{1}{2})\omega)}{\sin(\omega/2)}\varphi(\omega) \, d\omega.$$

Splitting the domain of integration according to $2\pi$-periodicity,

\begin{align*} T_M(\varphi) &= \sum_{k\in\mathbb{Z}} \ \mathrm{P.V.}\int_{-\pi}^{\pi} \frac{\cos((M+\frac{1}{2})(\omega+2\pi k))}{\sin((\omega+2\pi k)/2)}\varphi(\omega+2\pi k) \, d\omega \\ &= \sum_{k \in \mathbb{Z}} \int_{-\pi}^{\pi} \left( \frac{\cos((M+\frac{1}{2})\omega)}{\sin(\omega/2)}\varphi(\omega+2\pi k) - \frac{\varphi(2\pi k)}{\omega/2} \right) \, d\omega \end{align*}

Now writing

\begin{align*} &\int_{-\pi}^{\pi} \left( \frac{\cos((M+\frac{1}{2})\omega)}{\sin(\omega/2)}\varphi(\omega+2\pi k) - \frac{\varphi(2\pi k)}{\omega/2} \right) \, d\omega \\ &\hspace{4em} = \int_{-\pi}^{\pi} \left( \frac{\varphi(\omega+2\pi k)}{\sin(\omega/2)} - \frac{\varphi(2\pi k)}{\omega/2} \right) \cos((M+\tfrac{1}{2})\omega) \, d\omega \\ &\hspace{5em} + \varphi(2\pi k) \underbrace{\int_{-\pi}^{\pi} \frac{\cos((M+\tfrac{1}{2})\omega) - 1}{\omega/2} \, d\omega}_{=0\text{ by parity}} \end{align*}

we find that

$$T_M(\varphi) = \int_{-\pi}^{\pi} \underbrace{\left[ \sum_{k\in\mathbb{Z}} \left( \frac{\varphi(\omega+2\pi k)}{\sin(\omega/2)} - \frac{\varphi(2\pi k)}{\omega/2} \right) \right]}_{\in C([-\pi,\pi])} \cos((M+\tfrac{1}{2})\omega) \, d\omega$$

Therefore by the Riemann-Lebesgue lemma, we have $\lim_{n\to\infty} T_M(\varphi) = 0$ and hence $T_M \to 0$ in $\mathcal{D}'(\mathbb{R})$.

No, it is not true that the limit is zero for all $\omega$.

For example, if $\omega$ is irrational with respect to $\pi$, it follows from the Dirichclet Theorem on density that $\omega(M+1/2) \pmod{2 \pi}$ is dense in $[0, 2 \pi]$.

In particular, if $\frac{\omega}{\pi} \notin \mathbb Q$, it follows from the above that for each $\alpha \in [-1,1]$ there exists an increasing sequence $k_n$ of positive integers such that $$\lim_n \cos[\omega(k_n+1/2)] =\alpha$$

It follows that in this case the limit doesn't even exist.

Also, if $\frac{\omega}{\pi} \in \mathbb Q$, the sequence $\cos[\omega(M+1/2)]$ is periodic. Therefore, it is convergent if and only if it is constant, which happens only for one or two values of $\omega$.

• Maybe you are right, but the two arguments you show don't convince me. If the $\cos()$ in the numerator was replaced by a $\sin()$, I believe both of your arguments would apply, but the limit would equal $j\pi \delta(\omega)$, where $\delta$ is the Dirac delta function. Namely, the limit would be $0 \ \forall \omega\neq 0$, but there would be an impulse at $\omega =0$. Ergo, I don't think that your answer is enough to say that the limit in the OP isn't $0 \ \forall \omega$. – Tendero Jan 23 '18 at 1:07
• I just noticed that what I wrote above isn't exactly true. The limit would not be $0$ for every $\omega\neq 0$, because the impulses would be periodic. So if we replaced the cosine by a sine, the limit would actually equal $0 \ \forall \omega \neq 2k\pi$. – Tendero Jan 23 '18 at 1:19
• @Tendero No, even with $\sin$, the limit is NOT $j \pi \delta(\omega)$. It is true that with $\sin$, the sequence of FUNCTIONS converges to $j \pi \delta(\omega)$ in the WEAK-STAR topology on the space of distributions. But this is not what you are asking, and if this is what you are trying to ask, you need to make the question more precise. – N. S. Jan 23 '18 at 1:27
• Sorry about that, my knowledge in distribution theory is very shallow at the moment. I'll edit my question to make it clearer. – Tendero Jan 23 '18 at 1:29
• @Tendero But BIG WARNINg coming. Whenever when we say that $\lim_n f_n =\delta$ as distributions, this has absolutely nothing to do with the limit at some or many $x \in \mathbb R$. When we write $$\lim_n f_n =\delta$$ as distributions, the exact meaning is the following: for all $f$ functions which are infinitely many times differentiable and compactly supported we have $$\lim_n \int_{-\infty}^\infty f(t) f_n(t) dt = f(0)$$ – N. S. Jan 23 '18 at 1:51