# Norms of trigonometric polynomials

Jordan Bell
April 3, 2014
###### Theorem 1.

Let $1\leq p\leq q\leq\infty$. If $\hat{f}(j)=0$ for $|j|>n+1$ then

 $\|f\|_{q}\leq 5(n+1)^{\frac{1}{p}-\frac{1}{q}}\|f\|_{p}.$
###### Proof.

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

 $|K_{n}(t)|\leq\min\Big{(}n+1,\frac{\pi^{2}}{(n+1)t^{2}}\Big{)}.$

Then, for any $r\geq 1$,

 $\displaystyle\|K_{n}\|_{r}^{r}$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}|K_{n}(t)|^{r}dt$ $\displaystyle\leq$ $\displaystyle\frac{1}{2\pi}\int_{0}^{\frac{\pi}{n+1}}(n+1)^{r}dt+\frac{1}{2\pi% }\int_{\frac{\pi}{n+1}}^{2\pi}\Big{(}\frac{\pi^{2}}{(n+1)t^{2}}\Big{)}^{r}dt$ $\displaystyle=$ $\displaystyle\frac{(n+1)^{r-1}}{2}+\frac{1}{2}\frac{1}{(n+1)^{r}}\frac{1}{2r-1% }\Big{(}(n+1)^{2r-1}-\frac{1}{2^{2r-1}}\Big{)}$ $\displaystyle\leq$ $\displaystyle\frac{(n+1)^{r-1}}{2}+\frac{1}{2}\frac{1}{(n+1)^{r}}\frac{1}{2r-1% }(n+1)^{2r-1}$ $\displaystyle\leq$ $\displaystyle(n+1)^{r-1}.$

Hence $\|K_{n}\|_{r}\leq(n+1)^{1-\frac{1}{r}}$.

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

 $\|V_{n}\|_{r}\leq 2\|K_{2n+1}\|_{r}+\|K_{n}\|_{r}\leq 2(2n+2)^{1-\frac{1}{r}}+% (n+1)^{1-\frac{1}{r}}\leq 5(n+1)^{1-\frac{1}{r}}.$

For $|j|\leq n+1$ 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

 $\|f\|_{q}=\|V_{n}*f\|_{q}\leq\|V_{n}\|_{r}\|f\|_{p}\leq 5(n+1)^{\frac{1}{p}-% \frac{1}{q}}\|f\|_{p}.$

## References

• [1] Yitzhak Katznelson, An introduction to harmonic analysis, third ed., Cambridge Mathematical Library, Cambridge University Press, 2004.