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The Chebyshev alternation theorem says that $p^{\ast }\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f\in C[-1,1]$ if and only if there are at least $n+2$ ordered pointsand a sign $\epsilon \in \{-1,1\}$ such thatThus the extremal error has common magnitude and alternating sign at $n+2$ points.

For any $n\ge 0$, take $f_n=T_{n+1}$, the Chebyshev polynomial of degree $n+1$. It has $n+2$ alternating extrema of magnitude one on $[-1,1]$. The Chebyshev alternation theorem shows that the zero polynomial is best from ${\mathcal{P}}_n$, with error one. Since it also lies in ${\mathcal{P}}_{n-1}$ when $n\ge 1$,

Now suppose $f^{(n)}(x)>0$ throughout $[-1,1]$ and, contrary to the claim, $E_{n-1}(f)=E_n(f)$. A best $p\in {\mathcal{P}}_{n-1}$ would then also be best in ${\mathcal{P}}_n$. Its error $e=f-p$ would have $n+2$ alternating extrema by the alternation theorem, hence at least $n+1$ distinct zeros. Applying the Rolle theorem $n$ times gives a zero ofcontradicting positivity. Therefore

For $n\ge 1$, choose the integer $m$ withand define the partial sumIt belongs to ${\mathcal{P}}_n$, and the triangle inequality givesAt the $3^{m+1}+1$ pointsevery tail term has the same alternating sign becauseThusThere are at least $n+2$ such points because $n<3^{m+1}$. The Chebyshev alternation theorem provesFor $n=0$, the partial sum is empty and the same argument at $x=\pm 1$ gives $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$.

The trigonometric Chebyshev alternation theorem says that $p_n^{\ast }\in {\mathcal{T}}_n$ is best exactly when its error has at least $2n+2$ cyclically ordered extrema of equal magnitude and alternating sign. Let $5^m\le n<5^{m+1}$ and set $x_j=j\pi /5^{m+1}$. For every $k\ge m+1$,ThusThere are $2\cdot 5^{m+1}\ge 2n+2$ such extrema, while the triangle inequality bounds the tail by their common magnitude. Hence

For positive summable coefficients, the Weierstrass M-test gives a continuous uniform sum. The partial sum over $3^k\le n$ is the unique best uniform approximation of degree at most $n$. If $3^K$ is the first omitted frequency, every tail term equals $(-1{)}^j$ at $x_j=\cos (j\pi /3^K)$ because its frequency ratio is odd. The entire tail therefore attains alternating extrema equal in magnitude to its coefficient sum at more than $n+1$ points. The Chebyshev alternation theorem proves the assertion, including the empty partial sum at $n=0$.