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Let $E_{\ast }={\inf }_{\mathrm{deg}q\le n}\Vert f-q{\Vert }_{\infty }$, and choose a minimizing sequence $p_m$ with $\Vert f-p_m{\Vert }_{\infty }\to E_{\ast }$. Since $E_{\ast }\le \Vert f{\Vert }_{\infty }$, discarding finitely many terms ensures $\Vert p_m{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }+1$.

Fix $n+1$ distinct nodes $a_0,\dots ,a_n$. The Lagrange interpolation formula expressesThe vector of node values is bounded in ${\mathbb{R}}^{n+1}$, so the Bolzano-Weierstrass theorem supplies a subsequence converging to $(v_0,\dots ,v_n)$. Put $p=\sum v_j{\ell }_j$, of degree at most $n$. Because the ${\ell }_j$ are fixed bounded functions,along that subsequence. The reverse triangle inequality implies $\Vert f-p{\Vert }_{\infty }=E_{\ast }$. Thus a best uniform approximation exists, with no appeal to an unproved abstract existence result.

Here ${\mathcal{P}}_n$ is the vector space of real polynomials of degree at most $n$.

Write ${\Delta }_t^{2}f(x)=f(x+t)-2f(x)+f(x-t)$, and use the second modulus of smoothness ${\omega }_2(f,h)={\sup }_{|t|\le h}\Vert {\Delta }_t^{2}f{\Vert }_{\infty }$. The Jackson kernel is even, nonnegative and has unit integral. Pairing the two halves of its convolution therefore givesWe need a weighted moment estimate rather than just concentration of the Jackson kernel. For $|t|\le \pi$, the bounds $|\sin (t/2)|\ge |t|/\pi$ and $|\sin (nt/2)|\le \min (n|\sin (t/2)|,1)$ implyConsequently, splitting the integral at $1/n$ givesThe second modulus of smoothness satisfies ${\omega }_2(f,t)\le (1+t/h{)}^2{\omega }_2(f,h)$ for $t\ge 0$. To see this, choose $q=\lceil t/h\rceil$ when $t>0$, put $u=t/q$, and factor the function translation difference $T_t-I=(T_u-I){\sum }_{r=0}^{q-1}{T}_u^r$. Squaring gives at most $q^2$ translated second differences of step $u$. Each function translation is an isometry in the supremum norm, and centered and forward second differences have the same supremum norm. This proves the stated scaling bound.

Taking $h=1/n$, using $(1+n|t|{)}^2\le 2+2n^2{t}^2$, and integrating against the Jackson kernel now provesThus the Jackson operator estimate is uniform in both $f$ and $n$.

For the requested degree-$n$ best uniform approximation, note that the Jackson kernel, and hence $j_{m}f$, has trigonometric polynomial degree at most $2(m-1)$. Indeed the fourth power in the Jackson kernel is the squared modulus of a squared sum of $m$ consecutive complex exponentials. Choose $m=\lfloor n/2\rfloor +1$, so $2(m-1)\le n$ and $m\ge n/2$. The second-difference integral formula givesApplying the Jackson operator estimate with this $m$ yields the concise answerThe change of index is essential: $j_{n}f$ itself generally has degree $2n-2$, exceeding the permitted degree.

Use real-valued continuous functions and real trigonometric polynomials. The trigonometric Chebyshev alternation theorem says that $p\in {\mathcal{T}}_n$ is the unique best uniform approximation to $f$ if and only if there are distinct points, ordered within one period,Thus the required number of alternating extrema is $2n+2$. When the error is zero, the assertion is interpreted trivially and $p=f$ is unique. The count reflects $\dim {\mathcal{T}}_n=2n+1$ and the fact that a nonzero degree-$n$ trigonometric polynomial has at most $2n$ zeros in one period. In particular a strictly better candidate would force its difference from $p$ to have alternating signs at these points and hence at least $2n+2$ zeros, counting the closing interval around the circle. The trigonometric Chebyshev alternation theorem also gives necessity and uniqueness.

For any degree-at-most-$n$ trigonometric polynomial $t_n$, linearity and reproduction giveThe operator norm bound in the preceding part yieldsTake the infimum over all such trigonometric polynomials. By the definition of best uniform approximation,No choice of a minimizer is needed for this argument. It is an instance of the polynomial reproduction error bound: a bounded linear reproducing operator has error at most $1+\Vert P\Vert$ times the optimal error.

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.

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

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.

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 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 $n=0$.

For an odd integer $q\ge 3$ and positive summable coefficients $a_k$, the Weierstrass M-test makes $f(x)={\sum }_{k\ge 0}{a}_k\cos (q^{k}x)$ continuous. Its partial sum over $q^k\le n$ is the unique best uniform approximation by degree-at-most-$n$ trigonometric polynomials, for $n\ge 1$. If $Q$ is the first omitted frequency, at $x_j=j\pi /Q$ every omitted cosine equals $(-1{)}^j$ because all frequency ratios are odd. Thus the tail attains alternating extrema equal to its coefficient sum at $2Q\ge 2n+2$ points. The trigonometric Chebyshev alternation theorem proves the formula. For $a_k=r^{-k}$, $r>1$, the largest decay exponent is $\log r/\log q$.

The Remez algorithm searches for a best uniform approximation by solving alternation equations on a reference set, finding extrema of the error, and exchanging reference points. For real degree-at-most-$n$ polynomials, the equioscillation theorem characterizes success by $n+2$ ordered alternating maximal errors. The alternating-reference error bound supplies a certified lower bound at each step, while the largest absolute error supplies an upper bound. Their gap measures how close the current polynomial is to optimal, provided the upper bound is established globally rather than only on a finite grid.

The Runge phenomenon is the failure of increasing-degree polynomial interpolation at equally spaced interpolation nodes to converge to some smooth target functions, with large oscillations near the ends of the interval. It is a property of the approximation problem, and can occur even with exact arithmetic. A numerically stable evaluation method cannot remove it. Endpoint clustering, as in Chebyshev-Lobatto nodes, changes the approximation problem and can greatly improve it.^[ref]

Take $f(x)=1/(1+25x^2)$ on $[-1,1]$ and polynomial degree at most $10$. Use either $x_k=-1+k/5$ or $x_k=\cos (k\pi /10)$ for $k=0,\dots ,10$. Each interpolation polynomial passes through all eleven of its own samples, yet the first has severe endpoint oscillations and the second stays much closer to the target function. Evaluating both by the barycentric interpolation formula on $20,001$ equally spaced points, including the endpoints, gives sampled maximum absolute errors of approximately $1.915659$ and $0.132197$, respectively: about a $14.49$-fold reduction. These are maxima over the stated grid, not certified maxima over the continuous interval. The generator separately checks exact reproduction of all stored nodal values.

The polynomial interpolation error formula explains why interpolation nodes enter the analysis. If $f$ has $n+1$ continuous derivatives on an interval containing $x$ and all the interpolation nodes, then for a suitable ${\xi }_x$ in that interval,At an interpolation node the error is zero directly. Away from the interpolation nodes, both the product and the high-order derivative matter. A short proof applies Rolle's theorem repeatedly to $f(t)-p_n(t)-[f(x)-p_n(x)]{\prod }_i(t-x_i)/{\prod }_i(x-x_i)$, which vanishes at $n+2$ distinct points.^[ref]

The successful comparison here does not assert that Chebyshev-Lobatto nodes solve every approximation problem for every function. The derivatives, the polynomial degree, and measurement errors still matter. The Lebesgue constant of interpolation measures worst-case amplification of nodal errors, while the polynomial reproduction error bound relates interpolation to the best uniform approximation available at that polynomial degree.

For real $f\in C^{n+1}[0,1]$ with positive $(n+1)$st derivative, equality of consecutive best uniform approximation errors would make a best degree-at-most-$n$ polynomial also best at degree $n+1$. The Chebyshev alternation theorem then supplies $n+3$ alternating extrema of its nonzero error, hence $n+2$ distinct zeros. Repeated Rolle's theorem forces a zero of the $(n+1)$st derivative of that error, contradicting the assumed sign. The same conclusion holds for a strictly negative derivative by replacing $f$ with $-f$.

For real continuous $2\pi$-periodic $f$, a degree-at-most-$n$ real trigonometric polynomial $p$ is its unique best uniform approximation exactly when $f-p$ has $2n+2$ alternating extrema of magnitude $\Vert f-p{\Vert }_{\infty }$ at distinct ordered points in one period. The closing interval around the circle also counts. A strictly improving approximant would force its difference from $p$ to have at least $2n+2$ zeros, whereas a nonzero degree-$n$ trigonometric polynomial has at most $2n$.