# $L^{p}$ norms of a sine sum

Jordan Bell
April 3, 2014

George Pólya and Gábor Szegő, Problems and theorems in analysis, Volume II, volume 216 of Die Grundlehren der mathematischen Wissenschaften, Springer, 1976, translated from the German by C. E. Billigheimer.

The following is an estimate of the $L^{p}$ norms of a sum with nonnegative terms (p. 77, no. 38).

Let $\Gamma_{n}(t)=\sum_{k=1}^{n}\frac{|\sin kt|}{k}$. As $|\sin kt|\leq 1$, we have $\Gamma_{n}(t)\leq\sum_{k=1}^{n}\frac{1}{k}$, but we can give a sharper upper bound for $\Gamma_{n}(t)$ using the following two results. First, if $B_{n}(t)=\sum_{k=1}^{n}\frac{\cos kt}{k}$, then $B_{n}(t)\geq-1$ for all $t$ (p. 75, no. 28). Second , for all $t$ (p. 76, no. 34),

 $|\sin t|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{j=1}^{\infty}\frac{\cos 2jt}{4j^{2}-% 1}.$

Therefore

 $\displaystyle\Gamma_{n}(t)$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{n}\frac{1}{k}\left(\frac{2}{\pi}-\frac{4}{\pi}\sum_{j% =1}^{\infty}\frac{\cos 2jkt}{4j^{2}-1}\right)$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}-\frac{4}{\pi}\sum_{j=1}^{% \infty}\frac{1}{4j^{2}-1}\sum_{k=1}^{n}\frac{\cos 2jkt}{k}$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}-\frac{4}{\pi}\sum_{j=1}^{% \infty}\frac{B_{n}(2jt)}{4j^{2}-1}$ $\displaystyle\leq$ $\displaystyle\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}+\frac{4}{\pi}\sum_{j=1}^{% \infty}\frac{1}{4j^{2}-1}$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}+\frac{2}{\pi},$

using

 $\sum_{j=1}^{\infty}\frac{1}{4j^{2}-1}=\sum_{j=1}^{\infty}\frac{1}{2}\cdot\frac% {1}{2j-1}-\frac{1}{2}\cdot\frac{1}{2j+1}=\frac{1}{2}.$

Thus $\|\Gamma_{n}\|_{p}\leq\|\Gamma_{n}\|_{\infty}\leq\frac{2}{\pi}\sum_{k=1}^{n}% \frac{1}{k}+\frac{2}{\pi}$. On the other hand,

 $\displaystyle\|\Gamma_{n}\|_{1}$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\Gamma_{n}(t)dt$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k}\int_{0}^{2\pi}|\sin kt|dt$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k^{2}}\int_{0}^{2\pi k}|\sin t% |dt$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k}\int_{0}^{2\pi}|\sin t|dt$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k}\cdot 4$ $\displaystyle=$ $\displaystyle\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}.$

Therefore $\|\Gamma_{n}\|_{p}\geq\|\Gamma_{n}\|_{1}=\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}$.

Hence for $1\leq p\leq\infty$,

 $\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}\leq\|\Gamma_{n}\|_{p}\leq\frac{2}{\pi}+% \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}.$

So,

 $\|\Gamma_{n}\|_{p}=\frac{2}{\pi}\log n+O(1).$