= Solution
Write $\Delta_t^2f(x)=f(x+t)-2f(x)+f(x-t)$, and use the <second modulus of smoothness> $\omega_2(f,h)=\sup_{|t|\le h}\|\Delta_t^2f\|_\infty$. The <Jackson kernel> is even, nonnegative and has unit integral. Pairing the two halves of its <convolution> therefore gives
$$
j_nf(x)-f(x)=\frac12\int_{-\pi}^{\pi}\Delta_t^2f(x)J_n(t)\,dt.
$$
We need a weighted moment estimate rather than just concentration of the <Jackson kernel>. For $|t|\le\pi$, the bounds $|\sin(t/2)|\ge |t|/\pi$ and $|\sin(nt/2)|\le\min(n|\sin(t/2)|,1)$ imply
$$
0\le J_n(t)\le C\min\{n,n^{-3}|t|^{-4}\}.
$$
Consequently, splitting the integral at $1/n$ gives
$$
\int_{-\pi}^{\pi}t^2J_n(t)\,dt
\le 2C\left(n\int_0^{1/n}t^2\,dt+n^{-3}\int_{1/n}^{\pi}t^{-2}\,dt\right)\le C'n^{-2}.
$$
The <second modulus of smoothness> satisfies $\omega_2(f,t)\le(1+t/h)^2\omega_2(f,h)$ for $t\ge0$. To see this, choose $q=\lceil t/h\rceil$ when $t>0$, put $u=t/q$, and factor the <function translation> difference $T_t-I=(T_u-I)\sum_{r=0}^{q-1}T_u^r$. Squaring gives at most $q^2$ translated second differences of step $u$. Each <function translation> is an <isometry> in the <supremum norm>, and centered and forward second differences have the same <supremum norm>. This proves the stated scaling bound.
Taking $h=1/n$, using $(1+n|t|)^2\le2+2n^2t^2$, and integrating against the <Jackson kernel> now proves
$$
\|j_nf-f\|_\infty\le\frac12\omega_2(f,1/n)\int_{-\pi}^{\pi}(1+n|t|)^2J_n(t)\,dt\le C''\omega_2(f,1/n).
$$
Thus \b[the Jackson operator estimate is uniform in both $f$ and $n$].
For the requested degree-$n$ <best uniform approximation>, note that the <Jackson kernel>, and hence $j_mf$, has <trigonometric polynomial> degree at most $2(m-1)$. Indeed the fourth power in the <Jackson kernel> is the squared modulus of a squared sum of $m$ consecutive complex exponentials. Choose $m=\lfloor n/2\rfloor+1$, so $2(m-1)\le n$ and $m\ge n/2$. The <second-difference integral formula> gives
$$
\Delta_t^2f(x)=\int_{-|t|}^{|t|}(|t|-|u|)f''(x+u)\,du,
\qquad \omega_2(f,h)\le h^2\|f''\|_\infty.
$$
Applying the <Jackson operator estimate> with this $m$ yields the concise answer
$$
\boxed{E_n(f)\le \|j_mf-f\|_\infty\le \frac{4C''}{n^2}\|f''\|_\infty\quad(n\ge1).}
$$
The change of index is essential: $j_nf$ itself generally has degree $2n-2$, exceeding the permitted degree.
Back to article page
