= Solution
The <first Jackson theorem for periodic approximation> asserts that a universal constant $C$ satisfies
$$
E_n^{\mathrm{trig}}(g)\le C\,\omega(g,1/n),\qquad n\ge1,
$$
for every continuous $2\pi$-periodic function $g$, where the infimum is over degree-at-most-$n$ <trigonometric polynomials>.
Apply it to the even function $g(\theta)=f(\cos\theta)$. If $q_n$ is a <trigonometric polynomial> approximating $g$, its even part
$$
q_n^{\mathrm{ev}}(\theta)=\frac{q_n(\theta)+q_n(-\theta)}2
$$
has no larger error, because $g(-\theta)=g(\theta)$ and the triangle inequality bounds each averaged error by $\|g-q_n\|_\infty$. Every even <trigonometric polynomial> has the form
$$
q_n^{\mathrm{ev}}(\theta)=\sum_{j=0}^n b_j\cos(j\theta)
=p_n(\cos\theta),\qquad
p_n(x)=\sum_{j=0}^n b_jT_j(x).
$$
Here the <Chebyshev polynomials> $T_j$ have algebraic degree $j$, so $p_n$ has degree at most $n$. Surjectivity of cosine gives
$$
\|f-p_n\|_{C[-1,1]}=\|g-q_n^{\mathrm{ev}}\|_{C(\mathbb T)}.
$$
Conversely, every algebraic <polynomial> of degree at most $n$ yields such an even <trigonometric polynomial>. Taking infima therefore proves the exact identity $E_n^{\mathrm{alg}}(f)=E_n^{\mathrm{trig}}(\widetilde f)$, not merely an inequality. Combine this with the periodic theorem and the preceding <modulus of continuity> estimate:
$$
\boxed{E_n^{\mathrm{alg}}(f)\le C\,\omega(\widetilde f,1/n)
\le C\,\omega(f,1/n)}.
$$
Symmetrization, the degree correspondence and the equality of norms justify every step in transferring the <Jackson-type estimate>.
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