 Solution

Use the uniform norm and the central second difference ${\delta }_h^{2}f(x)=f(x+h)-2f(x)+f(x-h)$, withThe Jackson kernel is even, nonnegative and normalized. Pairing $t$ and $-t$ in its integral therefore givesWe will estimate this expression directly in terms of the second modulus of smoothness.

First prove the modulus scaling bound. With translations $T_{h}f(x)=f(x+h)$, the forward difference obeysEach function translation preserves the uniform norm, so $\Vert {\delta }_{mh}^{2}f{\Vert }_{\infty }\le m^2\Vert {\delta }_h^{2}f{\Vert }_{\infty }$. For $t\ne 0$, take $m=\lceil n|t|\rceil$ and $h=t/m$. Then $|h|\le 1/n$ andThe same bound is immediate at $t=0$.

For $|t|\le \pi$, $|\sin (t/2)|\ge |t|/\pi$. Also $|\sin (nt/2)/\sin (t/2)|\le n$, as follows by expressing the ratio, up to a unit-modulus factor, as a sum of $n$ complex exponentials. The given normalizing coefficient is at most a constant times $n^{-3}$. Consequentlyand these two estimates combine intoIndeed splitting at $n|t|=1$ shows this with an absolute change of constant. HenceSubstitute the modulus bound in the symmetrized convolution and take the supremum over $x$ to obtain the Jackson operator estimateThe constant is independent of $f$ and the positive integer $n$; the kernel's quartic decay is what makes the weighted second-difference integral uniformly bounded.