= Solution
Define the <second modulus of smoothness> by $\omega_2(f,h)=\sup_{|u|\leq h}\|f(\cdot+u)-2f+f(\cdot-u)\|_\infty$. The <Jackson kernel> is even, nonnegative and has integral one. Symmetrizing its <convolution> therefore gives
$$
j_n(f,x)-f(x)=\frac12\int_{-\pi}^{\pi}[f(x+t)-2f(x)+f(x-t)]J_n(t)\,dt.
$$
We 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 $\|(T_u^m-I)^2f\|\leq m^2\|(T_u-I)^2f\|$. Central and forward second differences have the same norm. For every $|v|\leq|t|$, take $m=\lceil |v|/h\rceil$ and $u=v/m$. Taking the supremum over these $v$ proves
$$
\omega_2(f,|t|)\leq(1+|t|/h)^2\omega_2(f,h).
$$
The zero-step case is immediate.
For $|t|\leq\pi$, the inequalities $|\sin(nt/2)|\leq n|\sin(t/2)|$, $|\sin(nt/2)|\leq1$, and $|\sin(t/2)|\geq |t|/\pi$ imply
$$
0\leq J_n(t)\leq C\min\left(n,\frac1{n^3t^4}\right).
$$
Use the first estimate on $|t|\leq1/n$ and the second outside it. Then
$$
\int_{-\pi}^{\pi}(1+n|t|)^2J_n(t)\,dt
\leq C\left[n\int_0^{1/n}1\,dt+\frac1n\int_{1/n}^{\pi}\frac{dt}{t^2}\right]\leq C'.
$$
All constants are independent of $n$ and $f$. Combining these estimates proves \b[the <Jackson operator estimate>]
$$
\boxed{\|j_n(f)-f\|_\infty\leq C\,\omega_2(f,1/n).}
$$
For $f\in C^2(\mathbb T)$, two applications of the <fundamental theorem of calculus> give the <second-difference integral formula>
$$
f(x+h)-2f(x)+f(x-h)=\int_{-h}^{h}(h-|u|)f''(x+u)\,du
$$
for $h\geq0$. Its absolute value is at most $h^2\|f''\|_\infty$, so $\omega_2(f,h)\leq h^2\|f''\|_\infty$.
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 $m\geq n/2$ for $n\geq1$, \b[the smooth-function approximation rate is]
$$
\boxed{E_n(f)\leq\|j_m(f)-f\|_\infty\leq\frac{C_1}{n^2}\|f''\|_\infty.}
$$
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