all derivatives of sinc are bounded by $1$

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

First of all, let us out that $\mathrm{sinc}(t)\le 1$ isbounded by the Jordan’s inequality. To the derivatives, letus write $\mathrm{sinc}$ as a Fourier integral,

$$\mathrm{sinc}(t)=\frac{1}{2}{\int }_{-1}^1{e}^{ixt}𝑑x.$$

Let $k=1,2,\mathrm{\dots }$. Then

$$\frac{d^k}{d{t}^k}\mathrm{sinc}(t)=\frac{1}{2}{\int }_{-1}^1{(ix)}^k{e}^{ixt}𝑑x.$$

and

$\left|{\displaystyle \frac{d^k}{d{t}^k}}\mathrm{sinc}(t)\right|$$=$$\left|{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-1}^1}{(ix)}^k{e}^{ixt}𝑑x\right|$

$\le$${\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-1}^1}|{(ix)}^k{e}^{ixt}|𝑑x$

$\le$${\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-1}^1}{|x|}^k𝑑x$

$\le$${\displaystyle \frac{1}{2}}\cdot 2{\displaystyle {\int }_0^1}{|x|}^k𝑑x$

$\le$${\displaystyle {\int }_0^1}{x}^k𝑑x$

$\le$${\displaystyle \frac{1}{k+1}}$

$1.$