$L_p$ norm of the Dirichlet Kernel

Question

I am trying to show that $$ \left(\int_{-1/2}^{1/2} \left|\dfrac{\sin(\pi(2N+1)x)}{\sin(\pi x)}\right|^p dx \right)^{1/p} \approx N^{\frac{p-1}{p}}$$ for all $1<p\leq \infty$ by using the fact that $|\sin(x)| \approx |x|$ if $x \in [-1/2, 1/2)$.

I have tried using the change $y=Nx$ and I end having the result $2N+1$. I do not know what I am doing wrong.

I would appreciate any help.

Answer 1

Let's look at the case $p = \infty$ first. That case is different from the others, so we must treat it separately. From the definition of the Dirichlet kernel as a sum of exponentials, it is clear that $D_N$ is a continuous (periodic) function, and $\lvert D_N(x)\rvert \leqslant 2N+1 = D_N(0)$. This immediately yields $\lVert D_N\rVert_{L^\infty} = 2N+1$.

For $1 < p < \infty$, we want to evaluate (not exactly, only the asymptotic behaviour) the integral

$$I_{p,N} := \int_0^{1/2} \biggl\lvert \frac{\sin \bigl(\pi(2N+1)x\bigr)}{\sin (\pi x)}\biggr\rvert^p \,dx,$$

the rest is multiplying with a constant ($2$) and taking the $p^{\text{th}}$ root. Substitution $u = \pi x$ gives another constant factor ($1/\pi$), which we can also ignore, at least for the moment. Thus we want to determine

$$A_{p,C} := \int_0^{\pi/2} \biggl\lvert \frac{\sin (Cx)}{\sin x}\biggr\rvert^p\,dx.$$

We are only interested in the particular values $C = 2N+1$, but $C$ is shorter, and it doesn't matter whether the constant is an integer or not for the calculations. As a first step, we write

$$A_{p,C} = \underbrace{\int_0^{\pi/2} \frac{\lvert \sin (Cx)\rvert^p}{x^p}\,dx}_{B_{p,C}} + \underbrace{\int_0^{\pi/2} \lvert \sin (Cx)\rvert^p\biggl(\frac{1}{(\sin x)^p} - \frac{1}{x^p}\biggr)dx}_{E_{p,C}}.$$

We bound the error term $E_{p,C}$ first. Since $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4)$, we see $\frac{x}{\sin x} = 1 + \frac{x^2}{6} + O(x^4)$, and $(1+y)^p = 1 + py + O(y^2)$ then yields $\bigl(\frac{x}{\sin x}\bigr)^p - 1 = \frac{p}{6} x^2 + O(x^4)$. We thus have

$$0 \leqslant E_{p,C} \leqslant b\int_0^{\pi/2} \lvert \sin (Cx)\rvert^p x^{2-p}\,dx$$

for a suitable constant $b \geqslant p/6$. For $p < 3$, $\lvert \sin (Cx)\rvert \leqslant 1$ gives an upper bound independent of $C$. For $p \geqslant 3$, we split the integral at $\frac{\pi}{2C}$ (assuming $C > 1$) and use $\lvert \sin (Cx)\rvert \leqslant Cx$ on the first part and $\lvert \sin (Cx)\rvert \leqslant 1$ on the second part, to obtain an upper bound of the form $K\cdot C^{p-3}$ for $p > 3$, and $K\cdot \log C$ for $p = 3$.

Now we look at the main part $B_{p,C}$. The substitution $y = Cx$ gives

$$C^{1-p}B_{p,C} = \int_0^{\frac{\pi C}{2}} \frac{\lvert \sin y\rvert^p}{y^p}\,dy = \underbrace{\int_0^\infty \frac{\lvert \sin y\rvert^p}{y^p}\,dy}_{K_p} - \int_{\frac{\pi C}{2}}^{\infty} \frac{\lvert \sin y\rvert^p}{y^p}\,dy.$$

With

$$\int_a^\infty \frac{\lvert\sin y\rvert^p}{y^p}\,dy \leqslant \int_a^\infty \frac{dy}{y^p} = \frac{1}{p-1}a^{1-p},$$

we find

$$B_{p,C} = K_pC^{p-1} + O(1),$$

and hence

$$A_{p,C} \sim K_pC^{p-1},$$

which gives

$$\lVert D_N\rVert_{L^p} \sim \tilde{K}_p \cdot (2N+1)^{\frac{p-1}{p}} \sim M_p\cdot N^{\frac{p-1}{p}}.$$

Answer 2

Here is an alternative and direct proof:
Throughout this posting $1<p<\infty$ is fixed. Let $I=\big(-\tfrac12,\tfrac12\big]$ and $I_N=\big(-\tfrac{1}{(2N+1)},\tfrac{1}{2N+2}\big]$. The endpoints of $I_N$ are the two solutions to $$D_N(x):=\frac{\sin((2N+1)\pi x)}{\sin\pi x}=0$$ closest to the origin. Recall that
$$\begin{matrix} \frac{2}{\pi}\leq\frac{\sin t}{t}&\text{for all}&0\leq t\leq\frac{\pi}{2}\\ \Big|\frac{\sin t}{t}\Big|\leq1&\text{for all}& t\in\mathbb{R} \end{matrix}$$

Over $I_N$, we have $$\begin{align} \int_{I_N}|D_N(x)|^p\,dx &=2\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\Big|^p\,dx\\ &\leq 2(\pi(2N+1)/2)^p\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\ &\leq 2\Big(\frac{\pi}{2}\Big)^p(2N+1)^{p-1} \end{align}$$ and $$\begin{align} \int_{I_N}|D_N(x)|^p\,dx &\geq 2(2N+1)^p\int^{\tfrac{1}{2N+1}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\ &\geq2(2N+1)^p\int^{\tfrac{1}{2(2N+1)}}_0\Big|\frac{\sin((2N+1)\pi x)}{(2N+1)\pi x}\Big|^p\,dx\\ &\geq\Big(\frac{2}{\pi}\Big)^p(2N+1)^{p-1} \end{align}$$

Over $I\setminus I_N$ we have $$\begin{align} \int_{I\setminus I_N}|D_N(x)|^p\,dx&=2\int^{\tfrac12}_{\tfrac{1}{2N+1}}\Big|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\Big|^p\,dx\\ &\leq2\big(\frac{\pi}{2}\big)^p\int^{\tfrac{1}{2}}_{\tfrac{1}{2N+1}}\Big|\frac{\sin((2N+1)\pi x)}{\pi x}\Big|^p\,dx\\ &=2\big(\frac{\pi}{2}\big)^p(2N+1)^{p-1} \int^{\tfrac{(2N+1)\pi}{2}}_{\pi}\Big|\frac{\sin y}{y}\Big|^p\,dy\\ &\leq 2\big(\frac{\pi}{2}\big)^p(2N+1)^{p-1}\int^\infty_{\pi}\Big|\frac{\sin y}{y}\Big|^p\,dy\\ &\leq \frac{\pi(2N+1)^{p-1}}{2^{p-1}(p-1)} \end{align} $$ Putting things together, we obtain $$ \frac{2}{\pi}(2N+1)^{1-\tfrac1p}\leq\|D_N\|_p\leq2^{1/p}\Big(\frac{\pi}{2}+\Big(\frac{\pi}{2(p-1)}\Big)^{1/p} \Big)(2N+1)^{1-\tfrac1p} $$ From this, we obtain constants $C_p,c_p>0$ such that $$c_pN^{1-\tfrac1p}\leq \|D_N\|_p\leq C_p N^{1-\tfrac1p}$$ for all $N$. This may not be the best constants however.