# all derivatives of sinc are bounded by $1$

Let us show that all derivatives of $\operatorname{sinc}$ are bounded by $1$.

First of all, let us out that $\operatorname{sinc}(t)\leq 1$ is bounded by the Jordan’s inequality  . To the derivatives, let us write $\operatorname{sinc}$ as a Fourier integral,

 $\operatorname{sinc}(t)={1\over 2}\int_{-1}^{1}e^{ixt}\,dx.$

Let $k=1,2,\ldots$. Then

 ${d^{k}\over dt^{k}}\operatorname{sinc}(t)={1\over 2}\int_{-1}^{1}(ix)^{k}e^{% ixt}\,dx.$

and

 $\displaystyle\left|{d^{k}\over dt^{k}}\operatorname{sinc}(t)\right|$ $\displaystyle=$ $\displaystyle\left|{1\over 2}\int_{-1}^{1}(ix)^{k}e^{ixt}\,dx\right|$ $\displaystyle\leq$ $\displaystyle{1\over 2}\int_{-1}^{1}|(ix)^{k}e^{ixt}|\,dx$ $\displaystyle\leq$ $\displaystyle{1\over 2}\int_{-1}^{1}|x|^{k}\,dx$ $\displaystyle\leq$ $\displaystyle{1\over 2}\cdot 2\int_{0}^{1}|x|^{k}\,dx$ $\displaystyle\leq$ $\displaystyle\int_{0}^{1}x^{k}\,dx$ $\displaystyle\leq$ $\displaystyle\frac{1}{k+1}$ $\displaystyle<$ $\displaystyle 1.$
Title all derivatives of sinc are bounded by $1$ AllDerivativesOfSincAreBoundedBy1 2013-03-22 15:39:03 2013-03-22 15:39:03 matte (1858) matte (1858) 10 matte (1858) Result msc 26A06