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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.

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.

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.

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