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If $N\in\mathbb{N}$, define the $N$-Dirichlet kernel as: $$D_N:(-\pi,\pi)\to\mathbb{R}, t\mapsto \sum_{n=-N}^{N} e^{int} = \frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\sin\left(\frac{t}{2}\right)}.$$

Does there exist $\delta\in(0,\pi)$ and $C>0$ such that for all $N\in\mathbb{N}$ $$\sup_{s\in[-\delta,\delta]} \left|\int_0^sD_N(t)\operatorname{d}t\right|\le C ?$$

I know that for every $\delta>0$ we have that $$ \int_0^\delta |D_N(t)|\operatorname{d}t \to +\infty, N\to+\infty ?$$ so, if the previous claim is true, it is due to cancellations between positive and negative terms.

Any help?

Bob
  • 5,705

1 Answers1

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If $s\in(-\pi,\pi)$ and $N\in\mathbb{N}$ then: $$\left|\int_0^s \frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\sin\left(\frac{t}{2}\right)}\operatorname{d}t\right|\le \left|\int_0^s \frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\frac{t}{2}}\operatorname{d}t\right|+\int_0^\pi \left| \frac{1}{\sin\left(\frac{t}{2}\right)}-\frac{1}{\frac{t}{2}}\right|\operatorname{d}t.$$ Now, it is clear using Taylor series that the second integral is a positive real number $C_2$ while, for the first integral, we have that: $$\left|\int_0^s \frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\frac{t}{2}}\operatorname{d}t\right|= 2\left|\int_0^{\left(N+\frac{1}{2}\right)s} \frac{\sin\left(u\right)}{u}\operatorname{d}u\right|\le 2\sup_{M\in\mathbb{R}}\left|\int_0^M \frac{\sin\left(u\right)}{u}\operatorname{d}u\right|.$$ Now, the function: $$\mathbb{R}\to\mathbb{R}, M\mapsto \int_0^M \frac{\sin\left(u\right)}{u}\operatorname{d}u$$ is bounded by some constant $C_1$, so: $$\forall s\in(-\pi,\pi), \forall N\in\mathbb{N}, \left|\int_0^s \frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\sin\left(\frac{t}{2}\right)}\operatorname{d}t\right|\\ \le 2\sup_{M\in\mathbb{R}}\left|\int_0^M \frac{\sin\left(u\right)}{u}\operatorname{d}u\right|+\int_0^\pi \left| \frac{1}{\sin\left(\frac{t}{2}\right)}-\frac{1}{\frac{t}{2}}\right|\operatorname{d}t\le 2 C_1+ C_2.$$

Bob
  • 5,705