Uniform approximation

Uniform approximation minimizes the supremum norm of the difference between a target function and an approximant.

The error is measured uniformly over the domain using the supremum norm. A sequence of approximants whose uniform error tends to zero converges by uniform convergence; the optimization problem and the convergence notion are distinct.

If interpolation at distinct sites $x_i$ in a finite-dimensional space of continuous functions is unique, its cardinal functions satisfy ${\ell }_i(x_j)={\delta }_{ij}$. The interpolation linear map is $Pf={\sum }_{i}f(x_i){\ell }_i$ and has operator norm ${\Lambda }_{\mathbf{x}}$ in the supremum norm. The upper bound is the triangle inequality. For the reverse bound choose node values to be the signs of the cardinal functions at a point maximizing their absolute sum, and extend those values by a continuous function of norm one. The polynomial reproduction error bound gives $\Vert f-Pf{\Vert }_{\infty }\le (1+{\Lambda }_{\mathbf{x}}){\inf }_{s\in \mathrm{range}P}\Vert f-s{\Vert }_{\infty }$.

For a real $n$-dimensional vector space of continuous functions on a compact interval, choose sites maximizing the absolute evaluation determinant in any basis. Linear independence ensures some nonzero evaluation determinant, so a maximizer exists and has distinct sites. Replacing row $i$ by evaluation at $t$ multiplies the determinant by the cardinal function ${\ell }_i(t)$. Maximality implies $|{\ell }_i(t)|\le 1$, hence the Lebesgue constant of interpolation is at most $n$. This gives a useful interpolation set without asserting that determinant maximization minimizes the Lebesgue constant of interpolation.

A Chebyshev system on a nondegenerate real interval is a family of $n+1$ real continuous functions such that every linear combination with coefficients not all zero has at most $n$ distinct zeros. The condition entails linear independence and unique interpolation at any $n+1$ distinct points. It is a zero-counting hypothesis, not a requirement to count multiplicities.

A Chebyshev system of real continuous functions on a circle has odd dimension. If its dimension is $d$, take $d$ cyclically ordered evaluation points and move them continuously to the next cyclic permutation without letting points collide. The evaluation determinant stays nonzero, so cannot change sign. The final cyclic row permutation multiplies it by $(-1{)}^{d-1}$; consequently $d$ must be odd. For example, $1,\cos x,\sin x,\dots ,\cos mx,\sin mx$ has dimension $2m+1$ and the required zero bound.

Exponentials with distinct real exponents form a Chebyshev system on every nondegenerate real interval. For an inductive proof, multiply a candidate combination by $e^{-{\lambda }_{0}x}$. If it has $n+1$ zeros, Rolle's theorem gives at least $n$ zeros of its derivative, a combination of the $n$ exponentials with exponents ${\lambda }_i-{\lambda }_0$, $i>0$. The inductive zero bound forces all those coefficients to vanish. The remaining constant also vanishes, a contradiction.

A family of $n+1$ real continuous functions is a Chebyshev system exactly when every evaluation matrix at $n+1$ distinct points is invertible. A singular evaluation matrix supplies a nonzero coefficient vector whose function vanishes at those points; conversely, a violation of the zero bound supplies such a vector in the kernel. This also gives the existence and uniqueness of interpolation from its span.

For an algebra $A\subseteq C(K)$ on a compact space, its uniform closure consists of functions that are uniform limits of sequences from $A$. It is a closed algebra in the uniform norm, because addition and multiplication preserve uniform convergence for bounded functions. If it contains every constant and separates points, the Stone-Weierstrass theorem makes the closure all of $C(K)$.

On $[a,b]\subset (0,1)$, the uniform closure of integer-coefficient polynomials contains $1/2$, since ${\sum }_{j=0}^{N}x(1-2x{)}^j\to 1/2$ uniformly. This closure is a ring, so it contains all dyadic constants and then all real constants. It contains $x$ and therefore every real-coefficient polynomial. The Stone-Weierstrass theorem makes it all of $C([a,b])$. On an interval including zero this fails, since the value of an integer-coefficient polynomial at zero is an integer.

Polynomial approximation represents a function by polynomials of bounded polynomial degree. On a compact real interval, the Weierstrass approximation theorem guarantees uniform approximation of every continuous function by such polynomials. The displayed best error quantifies the accuracy available at degree at most $n$.

For a positive-definite weighted inner product on a compact interval, a least-squares polynomial is the orthogonal projection of a function onto the finite-dimensional space ${\Pi }_n$ of polynomials of degree at most $n$. In a basis ${\varphi }_0,\dots ,{\varphi }_n$, its coefficients solve the normal equations $Gc=d$, where $G_{ij}=({\varphi }_i,{\varphi }_j)$ and $d_i=(u,{\varphi }_i)$. The Gram matrix is positive definite: $z^{T}Gz=\Vert \sum z_i{\varphi }_i{\Vert }^2>0$ for $z\ne 0$. Thus the coefficients exist and are unique. The residual is orthogonal to every candidate polynomial, giving $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$ and proving optimality. An orthogonal basis makes these equations diagonal, as in a least-squares polynomial in an orthogonal-polynomial basis.

For the weight $w(x)=x$ on $[0,1]$, the normal equations for projecting $\sqrt{1-x^2}$ onto $\mathrm{span}\{1,x\}$ have Gram matrix $\left(\begin{array}{cc}1/2& 1/3\\ 1/3& 1/4\end{array}\right)$ and right side $(1/3,\pi /16{)}^T$. Its determinant is $1/72$, and inversion gives the displayed polynomial. orthogonality of the residual to both basis functions proves best approximation in the weighted norm.

On a symmetric interval with an even positive weight, the best least-squares polynomial approximation to an odd function is odd, and the approximation to an even function is even. Indeed, the even and odd polynomial subspaces are orthogonal. For odd $u$, write a candidate as $p=p_o+p_e$. ThenDeleting a nonzero even component strictly reduces the error, so the minimizer has $p_e=0$. The same argument exchanges the parities for even $u$. This reduces the normal equations to the matching parity block and avoids computing coefficients known in advance to vanish.

If $n\ge 1$, $f$ is real and $C^n$ on an interval, and $f^{(n)}$ is nowhere zero there, the displayed inequality holds. If the errors were equal, a best degree-at-most-$n-1$ polynomial would also be best at degree $n$. The Chebyshev alternation theorem and intermediate value theorem would give $n+1$ zeros of the error. Applying Rolle's theorem $n$ times would force a zero of $f^{(n)}$, a contradiction. The zero-error case already contradicts $f^{(n)}\ne 0$. In particular this proves strict improvement for the exponential function at every positive degree step.

For a polynomial of degree at most $n$, this derivative estimate holds for $-1<x<1$. It controls the derivative away from the endpoints. Sampling it at Chebyshev polynomial roots supplies the hypotheses of the Chebyshev nodal derivative comparison, which controls the missing endpoint regions. The norm is taken on the entire interval, although the displayed pointwise bound concerns its interior.

Apply the Bernstein inequality for algebraic polynomials to the central interval $|x|\le \cos (\pi /(2n))$, using $\sin (\pi /(2n))\ge 1/n$. On the remaining regions apply the Chebyshev nodal derivative comparison to $p^{\prime }$. Since $T_n^{\prime }(\cos \theta )=n\sin (n\theta )/\sin \theta$, the identity expressing $\sin (n\theta )/\sin \theta$ as a sum of $n$ unit complex phases gives $|T_n^{\prime }|\le n^2$. The endpoint limit is $T_n^{\prime }(1)=n^2$, so the constant is sharp.

If a real degree-at-most-$n$ polynomial has absolute value at most one at every extremum $\cos (j\pi /n)$ of the Chebyshev polynomial $T_n$, its $k$th derivative on $[-1,1]$ is bounded by $T_n^{(k)}(1)$, for $1\le k\le n$. This strengthens the Markov inequality for polynomial derivatives by replacing a bound on the whole interval with these finitely many samples. The node set is rigid, as quantified by uniqueness of Chebyshev derivative-norming nodes.

For $n\ge 1$, any distinct node set in $[-1,1]$ different from the $n+1$ Chebyshev polynomial extrema admits one degree-$n$ polynomial $q$, bounded by one at all those nodes, whose every nonconstant derivative violates the corresponding sharp endpoint bound. In decreasing order set $q(t_i)=(-1{)}^i$. The endpoint derivative signs of Lagrange cardinal polynomials give $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$. Interpolating $T_n$ on the same nodes showsEach term is nonnegative and at least one is positive, because the other node set includes a nonextremal point. Uniqueness is a claim about node sets: permuting the same nodes does not change their norming property.

Suppose linear operators $T_n$ on a normed vector space satisfy $\Vert T_n\Vert \le C$ uniformly and $T_{n}p\to p$ on a dense subspace. The displayed triangle inequality for any element $p$ of that subspace first permits $p$ to approximate $f$, then lets $n$ increase, proving $T_{n}f\to f$ for every $f$. Applied to the Bernstein polynomial, the Weierstrass approximation theorem supplies the dense polynomial subspace and $C=1$.

Every real continuous function on a compact square can be represented as ${\sum }_{q=1}^5{G}_q(a_q(x)+b_q(y))$, with continuous one-variable functions and fixed inner functions $a_q,b_q$ independent of the represented function. Grid-separated additive coordinates give the inner functions. For a residual $r$ of norm $M$, assign $r$ at a sample point divided by three to each separated rectangle image, and continuously interpolate between images. Each point lies in at least three rectangles, so if $k\in \{3,4,5\}$ summands are accurate, the error is at most $[|1-k/3|+(5-k)/3]M+5\eta /3=2M/3+5\eta /3$. Choose oscillation $\eta \le M/10$ to obtain contraction $5/6$. Iteration gives outer functions by uniform convergence of corrections bounded by $M(5/6{)}^j/3$. Whole-plane reduction for continuous superposition removes compactness of the original domain without assuming boundedness of the original function.

For $f\in C({\mathbb{R}}^2)$, choose a positive continuous function $A$ with $A(t)\ge (1+t){\max }_{x^2+y^2\le t}|f(x,y)|$ for $t\ge 0$. Such an $A$ is obtained by piecewise linear interpolation of a sufficiently large increasing sequence. Then $b(x,y)=f(x,y)/A(x^2+y^2)$ tends uniformly to zero at infinity. Transport $b$ to $(0,1{)}^2$ by $x=\tan (\pi (s-1/2))$ and extend it by zero on the boundary; the extension is continuous. Apply the compact Kolmogorov-Arnold representation theorem, then multiply by $A(x^2+y^2)$. Multiplication introduces no additional primitive operation because $uv=[(u+v{)}^2-(u-v{)}^2]/4$ uses only addition and continuous one-variable squaring and scaling.

For five pairs of continuous functions on $[0,1]$, one can arrange that at every arbitrarily fine scale there are closed interval families whose product rectangles cover each point at least three times, and within each family the images under $a_q(x)+b_q(y)$ are mutually disjoint. To see this, put gaps near five shifted fine grids; gap sets for different shifts are disjoint. Each coordinate misses at most one family, so a pair misses at most two. Approximate any prescribed pair of functions by constants on the complementary intervals, interpolating across gaps. Perturb the finitely many constants to avoid all equalities between distinct pair sums. The rectangle images are then separated by a positive distance, a property preserved under sufficiently small uniform perturbations. For each required mesh size this property defines a dense open subset of $C([0,1]{)}^{10}$. The Baire category theorem gives a tuple satisfying every scale simultaneously.

For a compact convex set $K\subset \mathbb{C}$ and $\zeta \notin K$, strict separation gives a unit vector $u$ with $\mathrm{Re}(\bar{u}\zeta )>{\max }_K\mathrm{Re}(\bar{u}z)$. Taking $c=-Ru$ for large $R$ makes $q={\max }_K|z-c|/|\zeta -c|<1$. The polynomialhas uniform error at most $q^{N+1}/(|\zeta -c|(1-q))$. This elementary geometric series argument proves exterior-pole approximation without a general complex approximation theorem.

Let $z_0$ be a boundary point of a compact convex set and choose exterior points ${\zeta }_n\to z_0$. Approximate $(z-{\zeta }_n{)}^{-1}$ uniformly on the compact set by a polynomial with error at most $1/n$. At each fixed $z\ne z_0$, the reciprocal converges and so does the approximating polynomial. On a boundary accumulating at $z_0$, uniform convergence to the singular reciprocal is impossible because each polynomial is bounded there. This distinguishes pointwise convergence from uniform convergence without confusing the two norms.

The uniform approximation closure of the integer-coefficient polynomials on $[0,1]$ consists precisely of the continuous functions taking integer values at both endpoints. Necessity follows because evaluations at zero and one are integers and convergent integer sequences are eventually constant. For sufficiency, the rounded Bernstein polynomial has integer monomial coefficients and differs uniformly by a vanishing amount from the Bernstein polynomial, which converges uniformly to the target.

For a continuous periodic function setIt measures smoothness through second differences without requiring a derivative. For $h>0$, factoring the translation difference gives ${\omega }_2(f,t)\le (1+t/h{)}^2{\omega }_2(f,h)$. The second-difference integral formula gives ${\omega }_2(f,h)\le h^2\Vert f^{\prime \prime }{\Vert }_{\infty }$ for twice continuously differentiable functions.

For a twice continuously differentiable function and $h\ge 0$, the fundamental theorem of calculus givesThe nonnegative triangular weight has integral $h^2$, so its supremum norm estimate directly controls the second modulus of smoothness.

A uniformly continuous real function on a closed bounded interval can be approximated uniformly by finite step functions. Given $\epsilon >0$, choose a mesh smaller than the uniform-continuity distance and set the approximant equal to a sampled value on each interval. Then the supremum norm of the approximation error is at most $\epsilon$. If the function has a Lipschitz bound $L$, a mesh of maximum length $h$ gives the explicit error bound $\Vert f-g{\Vert }_{\infty }\le Lh$.

Given a family $A$ of approximating functions, a best uniform approximation to $f$ is an element $p\in A$ satisfying

For $0<k<1$, $x^k-x$ is nonnegative on $[0,1]$ and has its maximum $(1-k)k^{k/(1-k)}$ at $x=k^{1/(1-k)}$. Shifting the chord $x$ upwards by half this maximum creates three alternating errors of equal magnitude. The equioscillation theorem proves that the displayed line is the unique best uniform linear approximation, with error equal to its constant shift.

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.

Suppose a reference polynomial $q$ has nonzero alternating errors of magnitude $|h|$, and another point $\eta$ has error magnitude $M>|h|$. Insert $\eta$ and delete a reference point so that the retained errors, in spatial order, still alternate. Solving the new alternation equations gives the displayed strict increase in reference height, with $w_j={\prod }_{k\ne j}(z_j-z_k{)}^{-1}$ on the new ordered set. To prove this, apply the order-$(n+1)$ divided difference to the old and new degree-at-most-$n$ polynomials. Both vanish, so the new signed height is the sum of the old errors times $w_j$, divided by ${\sum }_j(-1{)}^j{w}_j$. All these signed weight products have one sign. Its absolute value is therefore the positive weighted average of one value $M$ and $n+1$ values $|h|$.

For an interior $\eta$, replace the endpoint of its containing reference interval with the same error sign. If $\eta$ is exterior, replace the nearest endpoint when the signs agree; if they disagree, insert $\eta$ and drop the farthest endpoint. These rules preserve ordered alternation. The nearest point with the same sign need not be an adjacent point on an irregular grid, and an unqualified nearest-same-sign replacement does not ensure the strict increase.

Cosine substitution is an isometry from continuous functions on $[-1,1]$ to even continuous $2\pi$-periodic functions. It satisfies $\omega (\tilde{f},\delta )\le \omega (f,\delta )$ because cosine is Lipschitz continuous with constant one. Even degree-at-most-$n$ trigonometric polynomials correspond exactly to algebraic polynomials via $T_j(\cos \theta )=\cos (j\theta )$. Averaging a periodic approximant with its reflection cannot increase its error against an even target, so the algebraic and periodic best errors are equal.

If a bounded linear map $P$ reproduces every element of an approximating polynomial space $V$, then $f-Pf=(f-v)-P(f-v)$ for every $v\in V$. The triangle inequality and infimum over $v$ prove the displayed error bound. The same argument works for any linear approximating space, and applies in particular to de la Vallée Poussin sums.

For a fixed degree bound, two best approximating polynomials have a best approximating average. At each active point of that average their errors agree, hence the two polynomials agree. There must be at least two more active points than the degree: otherwise polynomial interpolation constructs a strictly improving sign-matching perturbation. The difference has too many zeros and must vanish. Finite-dimensional coefficient compactness also gives existence.

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.

A degree-at-most-$m$ polynomial is a best uniform approximation precisely when its error attains alternating extrema of equal magnitude at at least $m+2$ ordered points.

On $[0,1]$, the chord of $\sqrt{1-x}$ is $1-x$. Its nonnegative gap has maximum $1/4$ at $x=3/4$, as differentiation shows. Shifting the chord upward by $1/8$ gives error $-1/8,1/8,-1/8$ at $0,3/4,1$. The Chebyshev alternation theorem therefore proves that $9/8-x$ is the unique best linear uniform approximation, with error $1/8$. Endpoint differentiability is unnecessary.

Let a polynomial $q$ of degree at most $n$ have errors $f(x_j)-q(x_j)=(-1{)}^{j}h$ at $n+2$ ordered distinct points. Then $|h|$ bounds below the error $E_n(f)$ of the best uniform approximation. Otherwise an improved polynomial $p$ with error strictly below $|h|$ would make $p-q$ alternate strictly in sign at all those points, giving at least $n+1$ distinct roots of a polynomial by the intermediate value theorem. Its degree is at most $n$, a contradiction. Moreover, if $q$ is not optimal, $|h|<E_n(f)$. In the equality case, an optimal $p$ gives weak alternating signs of $p-q$; its vanishing order-$(n+1)$ divided difference has same-sign summands, forcing every nodal value to vanish and hence $p=q$.

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 $2n$.

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 $\log r/\log q$.

A standard Weierstrass function is a uniformly convergent lacunary cosine series, here parameterized by $a>1$ and an integer frequency multiplier $b>1$. Its continuity follows from the Weierstrass M-test. In the range $1<a<b$, its modulus of continuity has the fractional-power scale ${\delta }^{\log a/\log b}$ for the positive odd-frequency model; the critical case $a=b$ can have the larger scale $\delta \log (1/\delta )$. These examples distinguish small approximation error from ordinary Lipschitz continuity.

For $g(x)={\sum }_{k\ge 0}{5}^{-k}\cos (5^{k}x)$, all frequencies satisfy $5^k\equiv 1(\mathrm{mod}4)$. At $x=\pi /2$, the increment is $-{\sum }_{k\ge 0}{5}^{-k}\sin (5^{k}h)$. If $5^m\le 1/h<5^{m+1}$, the first $m+1$ terms have $0<5^{k}h\le 1$ and contribute at least $(m+1)h\sin 1$, while the remaining tail has magnitude at most $5^{-m}/4\le 5h/4$. This proves a lower modulus of continuity bound of order $h\log (1/h)$ for small $h$. The inverse theorem for trigonometric approximation supplies the matching upper bound because the positive lacunary trigonometric series has $E_n(g)\asymp n^{-1}$.

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 $n=0$.

Bernstein's lethargy theorem says that best approximation errors in nested approximation spaces can tend to zero arbitrarily slowly. For algebraic polynomials on $[-1,1]$, given any decreasing positive sequence ${\delta }_n\to 0$, there is a continuous function $f$ whose best degree-$n$ uniform-approximation error is at least ${\delta }_n$ for every $n$.

For a strictly decreasing positive sequence ${ϵ}_n\to 0$, put $a_0={ϵ}_0-{ϵ}_1$ and $a_k={ϵ}_{3^{k-1}}-{ϵ}_{3^k}$ for $k\ge 1$. These are positive with sum ${ϵ}_0$. The positive lacunary Chebyshev series $f_{ϵ}={\sum }_k{a}_k{T}_{3^k}$ has error ${ϵ}_0$ at degree zero and error ${ϵ}_{3^{K-1}}\ge {ϵ}_n$ whenever $3^{K-1}\le n<3^K$. This proves the lower-bound form of Bernstein's lethargy theorem with an explicit continuous function.

A Jackson-type estimate bounds an approximation error by a smoothness quantity such as a modulus of continuity, with the smoothness evaluated at a scale comparable to the reciprocal of the approximating degree.

For $p\ge 1$ and $f\in C^p({\mathbb{T}}^n)$, there is a real trigonometric polynomial with coordinate frequencies $|k_j|\le N$ whose uniform error is at most $C_{n,p}{N}^{-p}\Vert f{\Vert }_{C^p}$. One obtains the multivariable statement by applying bounded one-dimensional Jackson approximation operators successively in each coordinate and telescoping their errors. Each operator commutes with differentiation in the other coordinates, and its uniform operator norm is bounded independently of $N$; the total error is bounded by a constant times $N^{-p}{\sum }_j\Vert {\partial }_j^{p}f{\Vert }_{\infty }$.

Every continuous $2\pi$-periodic function admits degree-at-most-$n$ trigonometric polynomials with supremum norm error bounded by a universal constant times its modulus of continuity at scale $1/n$, for $n\ge 1$. This first-order Jackson-type estimate transfers to algebraic polynomials through cosine substitution for polynomial approximation.

The normalized even kernelis nonnegative, has integral one, and is a trigonometric polynomial of degree $2(n-1)$. Its mass is concentrated on a scale $1/n$: $J_n(t)\le C\min (n,n^{-3}|t{|}^{-4})$ for $|t|\le \pi$. This controls a weighted second moment and produces the Jackson operator estimate.

For $j_{n}f(x)={\int }_{-\pi }^{\pi }f(x-t)J_n(t)dt$, evenness and unit mass reduce the error to an integral of second differences. The Jackson kernel bound and the scaling inequality for the second modulus of smoothness yield $\Vert j_{n}f-f{\Vert }_{\infty }\le C{\omega }_2(f,1/n)$. Since the degree is $2(n-1)$, a degree-$N$ estimate uses $n=\lfloor N/2\rfloor +1$. For $f\in C^2$ this gives $E_N(f)\le C^{\prime }{N}^{-2}\Vert f^{\prime \prime }{\Vert }_{\infty }$.

For periodic uniform approximation,

Define $h(x)=|x|\log (e\pi /|x|)$ on $[-\pi ,\pi ]$, set $h(0)=0$, and extend periodically. Concavity and monotonicity on $[0,\pi ]$ give $\omega (h,\delta )=\delta \log (e\pi /\delta )$ for $0<\delta \le \pi$. Its cosine Fourier coefficients areOne integration by parts gives the formula; the remaining integral is $\log j+O(1)$ by the Dirichlet test. All these coefficients are nonpositive. The de la Vallée Poussin sum $V_n=2{\sigma }_{2n}-{\sigma }_n$ reproduces degree-at-most-$n$ trigonometric polynomials and has operator norm at most three. Hence $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and annihilates those polynomials. Its weights on coefficients are zero up to $n$, nonnegative thereafter, and one from $2n$ onwards, so $|L_n(h)|\ge {\sum }_{j\ge 2n}|a_j|\gtrsim \log n/n$. Thus $E_n(h)\ge |L_n(h)|/4\gtrsim \log n/n$; the first Jackson theorem for periodic approximation proves the matching upper bound. This cusp and the critical Weierstrass modulus have the same first-modulus order but different best-approximation rates.

The first Jackson theorem for periodic approximation applied to $|\sin \theta |$ gives algebraic best error $O(n^{-1})$ for $\sqrt{1-x^2}$. Its ordinary interval modulus of continuity at $1/n$ is at least $\sqrt{2/n-1/n^2}$. Substituting that error rate into the unmodified inverse theorem for trigonometric approximation would instead bound the modulus by $O(\log (n+1)/n)$, a contradiction. Algebraic inverse estimates must incorporate the endpoint compression of cosine substitution for polynomial approximation.

Uniform convergence of positive linear operators on $1,x,x^2$ implies uniform convergence on every continuous function on a compact interval.

For positive linear operators on continuous functions on $[0,1]$, put $e_j=U_n(x^j)-x^j$. Uniform continuity supplies the barriers $|f(x)-f(t)|\le \epsilon +\gamma (x-t{)}^2$. Applying positivity and evaluating at $t$ gives the displayed estimate because $U_n((x-t{)}^2)(t)=e_2(t)-2t{e}_1(t)+t^2{e}_0(t)$. The constant $\gamma$ is independent of $t$ and $n$. Thus uniform reproduction of $1,x,x^2$ proves Korovkin theorem convergence on every continuous function; exact reproduction forces the operator to be the identity.

Let $K$ be a compact Hausdorff space. A test subspace $H\subset C(K,\mathbb{R})$ containing a strictly positive function is sufficient for uniform convergence of positive linear operators on continuous functions if, for each $x\in K$, the only positive linear functional agreeing with point evaluation on $H$ is that evaluation itself. By the Riesz-Markov-Kakutani representation theorem, this is uniqueness of the positive measure with the prescribed moments. Boundedness from the positive test function and the Banach-Alaoglu theorem prove the assertion by compactness. When $1\in H$, nonnegative tests with a unique zero at each prescribed $x$ give a convenient sufficient condition.

For $2\pi$-periodic continuous functions, convergence of positive linear operators on continuous functions to the identity on $1,\sin x,\cos x$ implies uniform convergence on every function. Indeed $1-\cos (t-x)$ is nonnegative and has just one zero on the circle, so the compact Korovkin test space criterion applies.

A spline is a piecewise polynomial function whose polynomial pieces obey specified smoothness conditions at their joins, called spline knots. For example, a cubic spline is cubic on each interval and commonly has two continuous derivatives across interior knots. Spline approximation and spline interpolation use these finite-dimensional function spaces to represent data or approximate functions.

A linear spline is a continuous piecewise linear function on a specified spline knot sequence. For uniform integer spline knots, the hat basis in the displayed representation interpolates the data $s(i)=c_i$. The derivative is constant on each open span and can jump at a knot. Coinciding neighboring slopes remove a particular jump, but generic data give only $C^0$ regularity.

For nodes $a=x_0<\cdots <x_n=b$ and prescribed real node values, let $s$ be the linear spline joining them. Among all absolutely continuous functions $u$ with the same node values and square-integrable derivative, $s$ uniquely minimizes the Dirichlet energy ${\int }_a^b|u^{\prime }{|}^2$.

On each node interval, $s^{\prime }$ is constant and $u-s$ vanishes at both endpoints. The fundamental theorem of calculus therefore gives $\int s^{\prime }(u^{\prime }-s^{\prime })=s^{\prime }[(u-s)(x_{i+1})-(u-s)(x_i)]=0$. Expanding the square and summing proves the displayed energy decomposition. Its final term is nonnegative; equality forces $u^{\prime }=s^{\prime }$ almost everywhere and the endpoint values fix the additive constant, proving uniqueness. In particular, the minimum is ${\sum }_i(u(x_{i+1})-u(x_i){)}^2/(x_{i+1}-x_i)$. If all interior node values are zero and there is at least one interior node, only the two endpoint intervals contribute to this minimum.

A spline quasi-interpolant synthesizes a B-spline expansion from local bounded linear functionals, instead of solving a global interpolation system. Suppose each functional is bounded by $c_k\Vert f{\Vert }_{C[t_i,t_{i+k}]}$ and the operator reproduces every spline. On a knot cell $[t_j,t_{j+1}]$, only indices $j+1-k\le i\le j$ contribute. Their supports lie in $[t_{j+1-k},t_{j+k}]$, and subpartition of unity for B-splines proves local operator norm at most $c_k$. Reproduction of degree-$k-1$ polynomials and a local Taylor theorem remainder then give error at most $(1+c_k)(2k-1{)}^k{h}^k\Vert f^{(k)}{\Vert }_{\infty }/k!$, on the basic knot domain with the usual completed boundary basis.

For a spanning direction matrix $\Xi =({\xi }_1,\dots ,{\xi }_m)$ in ${\mathbb{R}}^d$, the box spline is the density of the pushforward of uniform measure on $[0,1{]}^m$ under $t\mapsto \Xi t$. Equivalently,Its support is the zonotope ${\sum }_j[0,1]{\xi }_j$. It is piecewise polynomial of total degree at most $m-d$. For integer directions, splitting each integration interval in half yields its binary refinement symbolup to the monomial specifying the choice of origin. If $r$ is the smallest number of directions whose removal leaves a nonspanning set, the usual box spline smoothness criterion gives $C^{r-2}$ regularity. Repeated directions increase smoothness while enlarging support.