Norms of trigonometric polynomials

Theorem 1.

Let $\mathrm{1}\mathrm{\le }p\mathrm{\le }q\mathrm{\le }\mathrm{\infty }$. If $\widehat{f}\mathit{}\mathrm{(}j\mathrm{)}\mathrm{=}\mathrm{0}$ for $\mathrm{|}j\mathrm{|}\mathrm{>}n\mathrm{+}\mathrm{1}$ then

$${\parallel f\parallel }_q\le 5{(n+1)}^{\frac{1}{p}-\frac{1}{q}}{\parallel f\parallel }_p.$$

Proof.

Let $K_n(t)={\sum }_{j=-n}^n\left(1-\frac{|j|}{n+1}\right)e^{ijt}$, the Fejér kernel. From this expression we get $|K_n(t)|\le K_n(0)=n+1$. It’s straightforward to show that $K_n(t)=\frac{1}{n+1}{\left(\frac{\sin \frac{n+1}{2}t}{\sin \frac{1}{2}t}\right)}^2$. Since $\sin \frac{t}{2}>\frac{t}{\pi }$ for , we get $|K_n(t)|\le \frac{{\pi }^2}{(n+1)t^2}$, and thus we obtain

$$|K_n(t)|\le \min (n+1,\frac{{\pi }^2}{(n+1)t^2}).$$

Then, for any $r\ge 1$,

${\parallel K_n\parallel }_r^r$ $=$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|K_n(t)|}^r𝑑t$

$\le$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{\frac{\pi }{n+1}}}{(n+1)}^r𝑑t+{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{\frac{\pi }{n+1}}^{2\pi }}{\left({\displaystyle \frac{{\pi }^2}{(n+1)t^2}}\right)}^r𝑑t$

$=$ ${\displaystyle \frac{{(n+1)}^{r-1}}{2}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{(n+1)}^r}}{\displaystyle \frac{1}{2r-1}}\left({(n+1)}^{2r-1}-{\displaystyle \frac{1}{2^{2r-1}}}\right)$

$\le$ ${\displaystyle \frac{{(n+1)}^{r-1}}{2}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{(n+1)}^r}}{\displaystyle \frac{1}{2r-1}}{(n+1)}^{2r-1}$

$\le$ ${(n+1)}^{r-1}.$

Hence ${\parallel K_n\parallel }_r\le {(n+1)}^{1-\frac{1}{r}}$.

Let $V_n(t)=2K_{2n+1}(t)-K_n(t)$, the de la Vallée Poussin kernel [1, p. 16]. Then

$${\parallel V_n\parallel }_r\le 2{\parallel K_{2n+1}\parallel }_r+{\parallel K_n\parallel }_r\le 2{(2n+2)}^{1-\frac{1}{r}}+{(n+1)}^{1-\frac{1}{r}}\le 5{(n+1)}^{1-\frac{1}{r}}.$$

For $|j|\le n+1$ we have $\widehat{V_n}(j)=1$, and one thus checks that $V_n*f=f$. Take $\frac{1}{q}+1=\frac{1}{p}+\frac{1}{r}$. By Young’s inequality we have

$${\parallel f\parallel }_q={\parallel V_n*f\parallel }_q\le {\parallel V_n\parallel }_r{\parallel f\parallel }_p\le 5{(n+1)}^{\frac{1}{p}-\frac{1}{q}}{\parallel f\parallel }_p.$$

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