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The normalized even kernelis nonnegative, has integral one, and is a trigonometric polynomial of degree $2(n-1)$. Its mass is concentrated on a scale $1/n$: $J_n(t)\le C\min (n,n^{-3}|t{|}^{-4})$ for $|t|\le \pi$. This controls a weighted second moment and produces the Jackson operator estimate.

Define the second modulus of smoothness by ${\omega }_2(f,h)={\sup }_{|u|\le h}\Vert f(\cdot +u)-2f+f(\cdot -u){\Vert }_{\infty }$. The Jackson kernel is even, nonnegative and has integral one. Symmetrizing its convolution therefore givesWe first establish the needed scaling rule. For the translation operator $T_u$, $(T_u^m-I)=(I+T_u+\cdots +T_u^{m-1})(T_u-I)$. Squaring gives $\Vert (T_u^m-I{)}^{2}f\Vert \le m^2\Vert (T_u-I{)}^{2}f\Vert$. Central and forward second differences have the same norm. For every $|v|\le |t|$, take $m=\lceil |v|/h\rceil$ and $u=v/m$. Taking the supremum over these $v$ provesThe zero-step case is immediate.

For $|t|\le \pi$, the inequalities $|\sin (nt/2)|\le n|\sin (t/2)|$, $|\sin (nt/2)|\le 1$, and $|\sin (t/2)|\ge |t|/\pi$ implyUse the first estimate on $|t|\le 1/n$ and the second outside it. ThenAll constants are independent of $n$ and $f$. Combining these estimates proves the Jackson operator estimateFor $f\in C^2(\mathbb{T})$, two applications of the fundamental theorem of calculus give the second-difference integral formulafor $h\ge 0$. Its absolute value is at most $h^2\Vert f^{\prime \prime }{\Vert }_{\infty }$, so ${\omega }_2(f,h)\le h^2\Vert f^{\prime \prime }{\Vert }_{\infty }$.

Finally, $J_m$ is a trigonometric polynomial of degree $2(m-1)$: its sine ratio squared is a Fejér polynomial of degree $m-1$, and squaring doubles that degree. Thus $j_m(f)$ has degree at most $2(m-1)$. To bound degree-$n$ best approximation, choose $m=\lfloor n/2\rfloor +1$, rather than using $j_n$ directly. Since $m\ge n/2$ for $n\ge 1$, the smooth-function approximation rate is