Solution
= Solution
Twice applying the <fundamental theorem of calculus> gives
$$
f(x+h)-2f(x)+f(x-h)=\int_0^h\int_{-s}^{s}f''(x+u)\,du\,ds,
$$
hence $\boxed{\omega_2(f,t)\leq t^2\lVert f''\rVert_\infty}$. Part (b) gives $\lVert\sigma_n(f)-f\rVert_\infty=O(n^{-1})$. This cannot be little-$o$ for every $C^2$ function: for $f_0(x)=\cos x$,
$$
\sigma_n(f_0)=(1-n^{-1})\cos x,\qquad
\boxed{\lVert\sigma_n(f_0)-f_0\rVert_\infty=n^{-1}}.
$$