The normalised sinc function

The normalised sinc function is $\mathrm{sinc}:\mathbb{R}\to \mathbb{C}$, given for $t\mathrm{\ne }0$ by the quotient and at the origin by continuity,

$\mathrm{sinc}(t):=\frac{\sin (\pi t)}{\pi t}(t\mathrm{\ne }0),\mathrm{sinc}(0):=1.$

The values displayed show that $\mathrm{sinc}$ is real-valued: the sine and the identity of The derivatives of sine and cosine are cosine and minus sine are real functions there, and the quotient of real numbers is real.

Well-definedness. At every $t\mathrm{\ne }0$ the numerator and denominator are defined and $\pi t\mathrm{\ne }0$, so the quotient of real numbers is defined (The complex exponential by its power series is needed only to fix the ambient convention in which $\pi$, $\sin$ and the real numbers are embedded in $\mathbb{C}$). The value at $t=0$ is assigned separately; it is the correct limit, ${\lim }_{t\to 0}\mathrm{sinc}(t)=1$, because $\pi t\to 0$ as $t\to 0$ and ${\lim }_{u\to 0}\sin (u)\mathrm{/}u=1$ (The limit of sin x divided by x at zero is one), the substitution $u=\pi t$ being the case of Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$, or $g$ avoids $L$ on a punctured neighbourhood of $c$ in which the inner function $\pi t$ does not take the value $0$ away from $t=0$; with this value $\mathrm{sinc}$ is continuous at $0$, and it is continuous at every $t\mathrm{\ne }0$ because $t\mapsto \sin (\pi t)$ is a composite of continuous functions (The derivatives of sine and cosine are cosine and minus sine gives differentiability, hence continuity, and A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs) and $t\mapsto 1\mathrm{/}(\pi t)$ is a quotient with nonvanishing denominator, so Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function gives the quotient.

Symmetry and integer values. Sine is odd (Parity and the Pythagorean identity for sine and cosine), so for $t\mathrm{\ne }0$, $\mathrm{sinc}(-t)=\sin (-\pi t)\mathrm{/}(-\pi t)=\sin (\pi t)\mathrm{/}(\pi t)=\mathrm{sinc}(t)$, and the same identity holds at $t=0$; thus $\mathrm{sinc}$ is even. For an integer $k\mathrm{\ne }0$, $\sin (\pi k)=0$ by the zero set of sine (The zero sets of sine and cosine and the least positive common period 2 pi), so $\mathrm{sinc}(k)=0$, while $\mathrm{sinc}(0)=1$. Hence $\mathrm{sinc}(k)=0$ for every nonzero integer $k$, and the kernel vanishes at every sampling point except its own.

Bounds. For every real $t$, $\mathrm{\mid }\mathrm{sinc}(t)\mathrm{\mid }\le 1$: at $t=0$ this is an equality, and for $t\mathrm{\ne }0$ the one-Lipschitz estimate $\mathrm{\mid }\sin u-\sin 0\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$ (Sine and cosine are $1$-Lipschitz on $\mathbb{R}$) with $u=\pi t$ and $\sin 0=0$ (The derivatives of sine and cosine are cosine and minus sine) gives $\mathrm{\mid }\sin (\pi t)\mathrm{\mid }\le \pi \mathrm{\mid }t\mathrm{\mid }$, which proves the bound after division by $\mathrm{\mid }\pi t\mathrm{\mid }$. For every $t\mathrm{\ne }0$ one also has $\mathrm{\mid }\mathrm{sinc}(t)\mathrm{\mid }\le 1\mathrm{/}(\pi \mathrm{\mid }t\mathrm{\mid })$, since $\mathrm{\mid }\sin (\pi t)\mathrm{\mid }\le 1$ (Parity and the Pythagorean identity for sine and cosine) and division by $\mathrm{\mid }\pi t\mathrm{\mid }$ gives this tail estimate.

This is the normalisation used by the sampling theorem on this page: the reconstruction series is $f(x)={\sum }_{k}f(hk)\mathrm{sinc}(x\mathrm{/}h-k)$, and the kernel vanishes at every sampling point except its own, as proved above. The Fourier identity $\mathrm{sinc}(t)={\int }_{-1\mathrm{/}2}^{1\mathrm{/}2}{e}^{-2\pi it\xi }d\xi$ is not asserted by this definition; it is proved directly where the sampling theorem consumes it, from the complex primitive of the exponential. The complex exponential convention $e^{2\pi it}=\exp (2\pi it)$ underlying that display is $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$.

No choice principle is used in this item.