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For real $f$, a candidate $u\in U\subset C(K,\mathbb{R})$, and active set $E=\{x:|f(x)-u(x)|=\Vert f-u{\Vert }_{\infty }\}$, the candidate is best precisely when every $v\in U$ has some $x\in E$ with $(f(x)-u(x))v(x)\le 0$. A direction with strictly matching error signs on the whole active set gives a uniformly improving small perturbation by compactness. Conversely one oppositely signed active value already prevents improvement. This criterion yields the Chebyshev alternation theorem.

We work in the usual real-valued setting of the Chebyshev alternation theorem. For $f\in C[-1,1]$ and an algebraic polynomial $p$ of degree at most $n$, the theorem says that $p$ is a best uniform approximation if and only if its error has $n+2$ ordered points with alternating maximal values:If $f=p$, the zero-error case is included directly.

Existence follows, for example, by taking a minimizing sequence: its supremum norms are bounded, all norms on the finite-dimensional polynomial space are equivalent, and a convergent coefficient subsequence attains the infimum. To prove uniqueness of best uniform polynomial approximation, let $p,q$ both attain the minimum error $E$. Their average $r=(p+q)/2$ has error at most $E$ by the triangle inequality and therefore exactly $E$ by minimality.

If $E=0$, both polynomials equal $f$ and are equal. Otherwise apply the Chebyshev alternation theorem to $r$. At every alternating extremal point, $f-r$ is either $E$ or $-E$. But it is the average of $f-p$ and $f-q$, each lying in $[-E,E]$. An average attains an endpoint of this interval only when both entries equal that endpoint. Hence $p(x_j)=q(x_j)$ at all $n+2$ points. The difference is a degree-at-most-$n$ polynomial with more than $n$ distinct zeros, so it is identically zero. The best approximating polynomial is unique.

The original PDF has indices $3^k$; the exponent is lost in the TeX transcription. This lacunary indexing is essential to the positive lacunary Chebyshev series argument below.

Let $K$ be the least nonnegative integer with $3^K>n$, so $K=0$ for $n=0$. DefineFor $K=0$ the sum is empty and means the zero polynomial. Each included Chebyshev polynomial has degree at most $n$. Also $|T_j(x)|\le 1$ on the interval, so summability of the positive coefficients gives uniform convergence by the Weierstrass M-test, andChoose $L=3^K$ and the $L+1$ points $x_j=\cos (j\pi /L)$, $0\le j\le L$. For every omitted index $k\ge K$, the integer $3^{k-K}$ is odd. Consequently,Every term of the tail has the same sign at a given point, and thereforeThis shows both that the error norm is exactly $A_K$ and that it alternates at $L+1\ge n+2$ distinct points. The points are in decreasing order; reversing their order still gives alternation. The Chebyshev alternation theorem proves that this partial sum is the unique best uniform approximation. HenceIn particular, $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$. The equal signs of all tail terms at the same extrema are the reason positivity and the odd integer frequency ratios are useful.

Let $p\in {\mathcal{P}}_n$ and $e=f-p$, with $d=\Vert e{\Vert }_{\infty }>0$. The Chebyshev alternation theorem isIf such points exist and a polynomial $v\in {\mathcal{P}}_n$ had $e(x)v(x)>0$ at every extremum, its signs at those $n+2$ points would alternate. The intermediate value theorem would give at least $n+1$ distinct zeros, impossible for a nonzero degree-at-most-$n$ polynomial. The Kolmogorov criterion for uniform approximation therefore proves sufficiency.

For necessity, let $E_{+}=\{e=d\}$ and $E_{-}=\{e=-d\}$. They are disjoint compact sets. If one set is empty there is just one sign block. Otherwise their positive separation implies that, reading the extremal set from left to right, its signs form finitely many alternating blocks. If there are fewer than $n+2$ alternating extrema, there are at most $n+1$ blocks. Place one point ${\xi }_j$ in each extremum-free gap between consecutive blocks. A scalar multiple of $v(x)={\prod }_j(x-{\xi }_j)$ can be chosen to have exactly the sign of $e$ on every block. Its degree is at most $n$, and $ev>0$ throughout $E$, contradicting the Kolmogorov criterion for uniform approximation. Thus there must be at least $n+2$ alternating extrema. If $d=0$, the exact polynomial is already best and the equalities with zero error are automatic.

The Chebyshev polynomial identity $T_m(\cos \theta )=\cos (m\theta )$ implies $|T_m(x)|\le 1$ on $[-1,1]$. Since ${\sum }_i{\gamma }_i<\infty$, the Weierstrass M-test shows thatconverges uniformly on $[-1,1]$. Every partial sum is continuous, so the uniform limit theorem makes $f$ a well-defined continuous function.

LetThen $-1=x_0<x_1<\cdots <x_{3^{n+1}}=1$. Since every $3^i$ is odd and, for $i\ge n+1$, $3^{i-n-1}$ is an odd integer,It follows thatThus the error has equal alternating extrema at $3^{n+1}+1$ ordered points. The Chebyshev alternation theorem shows that $P_n$ is a best uniform approximation among polynomials of degree at most $3^{n+1}-1$, and in particular

Now let $({\delta }_N)$ be the prescribed decreasing positive sequence. DefineThen $s_j$ decreases strictly to zero, so every ${\gamma }_j>0$, the series ${\sum }_j{\gamma }_j=s_0$ converges, andGiven $N\ge 1$, choose $j\ge 0$ with $3^j\le N\le 3^{j+1}-1$. Monotonicity of best-approximation errors and of ${\delta }_N$ givesHence this continuous $f$ satisfiesfor every polynomial $P$ of degree at most $N$. This is the polynomial form of Bernstein's lethargy theorem.

The Chebyshev alternation theorem says that a polynomial $P$ of degree at most $n$ is a best uniform approximation to $f\in C([0,1])$ if and only if there are $n+2$ pointsand a sign $\epsilon \in \{1,-1\}$ for whichThis is the equiripple criterion.

To prove sufficiency, put $E=\Vert f-P{\Vert }_{\infty }$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\Vert f-Q{\Vert }_{\infty }<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The intermediate value theorem then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The Lagrange root bound over a field forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\mathrm{deg}R\le n$. Therefore no $Q$ gives a smaller error, and $P$ is best.

As printed, this part contains an indexing error. The displayed product ends at $m-1$, but the hypotheses contain $m+1$ alternating intervals indexed by $0,\dots ,m$. On the first $m$ intervals the printed polynomial has sign $(-1{)}^i$, while on the last interval it has sign $(-1{)}^{m-1}$ rather than the required $(-1{)}^m$. For example, when $m=1$ the printed product is the constant $Q=1$, which cannot reduce a negative extremum of $g$ on the final interval. The intended polynomial isWe prove the stated conclusion with this correction.

Each root $w_j$ lies in the gap between the $(j-1)$st and $j$th active intervals. Therefore $Q$ has sign $(-1{)}^i$ throughout $[u_i,v_i]$. In particular, wherever $|g|=M$, the numbers $Q$ and $g$ have the same sign, and $Q$ is nonzero.

LetThese are compact sets. On their union, the continuous function $|Q|$ has a positive minimum. Hence, throughout some open neighbourhood $U$ of $K_{+}\cup K_{-}$, subtracting a sufficiently small positive multiple $\eta Q$ moves $g$ strictly towards zero and gives $|g-\eta Q|<M$. On the compact complement $[0,1]\setminus U$, continuity gives a uniform margin $|g|\le M-ϵ$ for some $ϵ>0$. Taking alsoshows that $|g-etaQ|<M$ there. Thusfor every sufficiently small $\eta >0$.

It remains to prove necessity in the Chebyshev alternation theorem. Let $P$ be a best approximation of degree at most $n$, put $g=f-P$, and let $M=\Vert g{\Vert }_{\infty }$. If $g$ has no sequence of $n+2$ extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into $m+1$ alternating compact groups with $m\le n$. Choose disjoint intervals $[u_i,v_i]$ containing those groups and separated by regions on which $|g|<M$. After replacing $g$ by $-g$ if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial $Q$ of degree $m\le n$ and a small $\eta >0$ such thatThis contradicts the assumed optimality of $P$. Therefore every best polynomial has the required $n+2$ alternating extrema.