= Solution
Use the <uniform norm> and the central second difference $\delta_h^2f(x)=f(x+h)-2f(x)+f(x-h)$, with
$$
\omega_2(f,a)=\sup_{|h|\leq a}\|\delta_h^2f\|_\infty.
$$
The <Jackson kernel> is even, nonnegative and normalized. Pairing $t$ and $-t$ in its integral therefore gives
$$
j_n f(x)-f(x)=\frac12\int_{-\pi}^{\pi}\delta_t^2f(x)J_n(t)\,dt.
$$
We will estimate this expression directly in terms of the <second modulus of smoothness>.
First prove the modulus scaling bound. With translations $T_hf(x)=f(x+h)$, the forward difference obeys
$$
(T_{mh}-I)^2=(T_h-I)^2\left(\sum_{r=0}^{m-1}T_{rh}\right)^2.
$$
Each <function translation> preserves the <uniform norm>, so $\|\delta_{mh}^2f\|_\infty\leq m^2\|\delta_h^2f\|_\infty$. For $t\ne0$, take $m=\lceil n|t|\rceil$ and $h=t/m$. Then $|h|\leq1/n$ and
$$
\|\delta_t^2f\|_\infty\leq(1+n|t|)^2\omega_2(f,1/n).
$$
The same bound is immediate at $t=0$.
For $|t|\leq\pi$, $|\sin(t/2)|\geq|t|/\pi$. Also $|\sin(nt/2)/\sin(t/2)|\leq n$, 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 $n^{-3}$. Consequently
$$
J_n(t)\leq Cn,\qquad
J_n(t)\leq\frac{C}{n^3|t|^4}\quad(t\ne0),
$$
and these two estimates combine into
$$
J_n(t)\leq\frac{C'n}{(1+n|t|)^4}.
$$
Indeed splitting at $n|t|=1$ shows this with an absolute change of constant. Hence
$$
\int_{-\pi}^{\pi}(1+n|t|)^2J_n(t)\,dt
\leq2C'\int_0^{n\pi}\frac{du}{(1+u)^2}\leq2C'.
$$
Substitute the modulus bound in the symmetrized <convolution> and take the supremum over $x$ to obtain the <Jackson operator estimate>
$$
\boxed{\|j_nf-f\|_\infty\leq C'\omega_2(f,1/n).}
$$
The 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.
Back to article page
