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The even positive Fejér kernel of total mass one gives ${\sigma }_{n}f-f=\frac{1}{2}\int F_n(t)[f(\cdot +t)-2f+f(\cdot -t)]dt$. The scaling property of the second modulus of smoothness bounds this by ${\omega }_2(f,\delta )(1+{\delta }^{-2}\int t^2{F}_n(t)dt)$. On $[-\pi ,\pi ]$, $\sin (|t|/2)\ge |t|/\pi$ gives $\int t^2{F}_n(t)dt\le {\pi }^2/n$. Choose $\delta =n^{-1/2}$. For twice continuously differentiable functions, the second-difference integral formula makes the error $O(n^{-1})$.

Use the central second difference ${\Delta }_t^{2}f(x)=f(x+t)-2f(x)+f(x-t)$ and define the second modulus of smoothness byWe first need its scaling inequality. If $q$ is a positive integer, translation by $h$ gives the telescoping identityThe weights sum to $q^2$. For $t>0$, choose $q=\lceil t/\delta \rceil$ and $h=t/q\le \delta$; thenThe same bound covers $t=0$ trivially.

In the paper's normalization, the Fejér kernel is even and nonnegative, with ${\pi }^{-1}{\int }_{-\pi }^{\pi }{F}_n=1$. Pairing the positive and negative translations expresses the Fejér sum error asUse the scaling bound and $(1+s{)}^2\le 2(1+s^2)$ to obtainFor $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$. Bounding the numerator sine by one givesThe bound at $t=0$ holds by continuity. Choosing $\delta =n^{-1/2}$ proves the Fejér second-modulus approximation bound with an explicit universal constant:Finally, for twice continuously differentiable periodic $f$, two applications of the fundamental theorem of calculus give the second-difference integral formulaThe triangular weight has integral $h^2$, so ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$. Therefore

Use the centered second modulus of smoothnessThe relevant modulus properties are ${\omega }_2(f,t)\le (1+t/\delta {)}^2{\omega }_2(f,\delta )$ for $t\ge 0$, $\delta >0$, and ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$ when $f^{\prime \prime }$ is continuous. The latter follows from the second-difference integral formulaThe Fejér kernel is even, nonnegative and has integral one. Averaging the convolution at $t$ and $-t$ therefore givesApply the second modulus of smoothness bound and $(1+s{)}^2\le 2(1+s^2)$:For $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, while ${\sin }^2(nt/2)\le 1$. ThusThe removable value at $t=0$ causes no problem. With $\delta =n^{-1/2}$, the Fejér second-modulus approximation bound isIf $f^{\prime \prime }$ is continuous, this also yields $\Vert {\sigma }_{n}f-f{\Vert }_{\infty }\le (1+{\pi }^2)\Vert f^{\prime \prime }{\Vert }_{\infty }/n=O(n^{-1})$.

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

For a continuous periodic function setIt measures smoothness through second differences without requiring a derivative. For $h>0$, factoring the translation difference gives ${\omega }_2(f,t)\le (1+t/h{)}^2{\omega }_2(f,h)$. The second-difference integral formula gives ${\omega }_2(f,h)\le h^2\Vert f^{\prime \prime }{\Vert }_{\infty }$ for twice continuously differentiable functions.