 Solution

Write ${\Delta }_t^{2}f(x)=f(x+t)-2f(x)+f(x-t)$, and use the second modulus of smoothness ${\omega }_2(f,h)={\sup }_{|t|\le h}\Vert {\Delta }_t^{2}f{\Vert }_{\infty }$. The Jackson kernel is even, nonnegative and has unit integral. Pairing the two halves of its convolution therefore givesWe 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 $|\sin (nt/2)|\le \min (n|\sin (t/2)|,1)$ implyConsequently, splitting the integral at $1/n$ givesThe second modulus of smoothness satisfies ${\omega }_2(f,t)\le (1+t/h{)}^2{\omega }_2(f,h)$ for $t\ge 0$. 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 $(1+n|t|{)}^2\le 2+2n^2{t}^2$, and integrating against the Jackson kernel now provesThus 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 $j_{m}f$, 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 givesApplying the Jackson operator estimate with this $m$ yields the concise answerThe change of index is essential: $j_{n}f$ itself generally has degree $2n-2$, exceeding the permitted degree.