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Gaussian quadrature is a quadrature rule that chooses both sample locations and coefficients to achieve unusually high polynomial exactness. For a finite interval $[a,b]$ with $a<b$, assume the integration weight $w$ is integrable and positive almost everywhere. DefineLet $p_n$ be the monic orthogonal polynomial of polynomial degree $n$ for the inner product $⟨u,v{⟩}_w={\int }_a^{b}u(x)v(x)w(x)dx$: it satisfies $⟨p_n,q{⟩}_w=0$ for every polynomial $q$ of polynomial degree below $n$. The polynomial roots of $p_n$ are distinct and inside $(a,b)$; take them as the nodes $x_i$. Integrate the Lagrange cardinal polynomials ${\ell }_i$ to obtain $A_i=I_w({\ell }_i)$. These coefficients are positive, andThe integration weight $w(x)$ and the discrete coefficients $A_i$ play different roles: $w$ specifies the integral being approximated, while $A_i$ tell us how much each sample contributes.^[ref]

A proof outline explains the gain over simply integrating an arbitrary polynomial interpolation. Use polynomial division to write $f=p_{n}q+r$, with $\mathrm{deg}q\le n-1$ and $\mathrm{deg}r\le n-1$. Orthogonality makes $I_w(p_{n}q)=0$, and sampling at the polynomial roots of $p_n$ makes $Q_n(p_{n}q)=0$. The Lagrange cardinal polynomials recover $r$ exactly, so both expressions agree. Also $A_i=Q_n({\ell }_i^2)=I_w({\ell }_i^2)>0$, since ${\ell }_i^2$ has polynomial degree $2n-2$. This is the mechanism behind positive weights of Gaussian quadrature.

For a concrete two-node example, set $w=1$ on $[-1,1]$. The Legendre polynomial $P_2(x)=(3x^2-1)/2$ has polynomial roots $x=\pm 1/\sqrt{3}$. Its monic polynomial version $p_2=x^2-1/3$ has zero integral against both $1$ and $x$, as required. Symmetry and exactness for the constant polynomial give $A_1=A_2=1$. Thus the Gauss–Legendre instance of Gaussian quadrature isDirect evaluation on $1,x,x^2,x^3$ gives $2,0,2/3,0$, exactly their integrals. For the polynomial $f(x)=1+x+x^2+x^3$, two samples therefore give $8/3$ exactly. But for $x^4$ the rule gives $2/9$, whereas the integral is $2/5$: the error $I_1-Q_2$ is $8/45$.

The ceiling $2n-1$ is sharp for $n$ distinct real nodes under the positive-weight assumptions. Their squared nodal polynomial ${\prod }_i(x-x_i{)}^2$ has polynomial degree $2n$, is zero at every sample, and has a strictly positive weighted integral. Thus no such quadrature rule integrates every polynomial of polynomial degree $2n$ exactly. Nor does Gaussian quadrature integrate every nonpolynomial function exactly: the same two-node example gives $Q_2(|x|)=2/\sqrt{3}$, while $I_1(|x|)=1$.

Gaussian quadrature is useful when evaluations are costly and a modest polynomial captures the integrand well. More precisely, for a continuous function $f$ and any polynomial $r$ of polynomial degree at most $2n-1$, positivity and constant exactness giveThe uniform norm on $[a,b]$ measures the approximation between samples. This bound connects quadrature rule accuracy to uniform approximation, rather than promising an error from sample count alone. Corners, sharp unresolved features and roundoff error require their own assessment; the absolute value function example already shows why polynomial exactness is a qualified guarantee.

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.

Suppose a family $A\subseteq C(K,\mathbb{R})$ on a compact space is closed under pointwise maximum and minimum. If a target $f$ can be approximated arbitrarily closely at each pair of points by a member of $A$, it can be approximated in the supremum norm. For a fixed first point, take finitely many pairwise approximants whose neighborhoods cover $K$ and take their maximum: it lies above $f-\epsilon$ everywhere and below $f+\epsilon$ near that first point. Take finitely many such first-point neighborhoods covering $K$ and take the minimum of their maxima. The result is within $\epsilon$ of $f$ everywhere. Starting with tolerance $\epsilon /2$ ensures a strict uniform approximation error smaller than $\epsilon$.

A unital Banach star-algebra is a complex unital Banach algebra with a continuous conjugate-linear algebra involution satisfyingThe algebra norm is complete and submultiplicative, with the unit conventionally normalized to norm one. Many definitions additionally require the involution to be isometric, $\Vert a^{\ast }\Vert =\Vert a\Vert$; the function algebra here satisfies this stronger convention as well. A Banach star-algebra does not by definition require the extra C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

Here is the lattice proof of the Stone-Weierstrass theorem in its complex self-adjoint form. Let $B$ be the uniform closure of the given unital star-subalgebra in $C(\Omega )$, and let $B_{\mathbb{R}}$ be its real-valued part. Closure preserves multiplication and complex conjugation, so $B$ is still a unital star-algebra. Real and imaginary parts lie in $B_{\mathbb{R}}$, and separation by the complex algebra implies separation by its real-valued part.

For $h\in B_{\mathbb{R}}$, polynomial approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb{R}}$. For example, this polynomial approximation follows from the binomial series for the square root, whose coefficients give uniform convergence on the relevant closed interval. Thusalso lie in $B_{\mathbb{R}}$. It is therefore a lattice of functions.

To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $ϵ>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb{R}}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.

Fix $x$. The sets where $h_{xy}>f-ϵ$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb{R}}$ with $g_x>f-ϵ$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+ϵ$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb{R}}$ satisfyingThis is lattice approximation from two-point approximation. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb{R}}$ is closed, it contains every real continuous function. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega )}$.

The annulus test. For the closed annulus, the family holomorphic in the interior is not dense in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so locally uniform convergence of holomorphic functions makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega )$. It is proper because $z\mapsto \overline{z}$ is continuous but not holomorphic on the annulus interior.

There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)dz$ around its positively oriented boundary. The Cauchy integral theorem gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline{z})=2\pi i{r}^2\ne 0$. Since $|L(h)|\le 2\pi r\Vert h{\Vert }_{\infty }$, no uniform approximation of $\overline{z}$ by that family is possible.

The original PDF prints $O(n^{\alpha })$, with a positive exponent. That premise is insufficient; the intended decay rate is $O(n^{-\alpha })$. Indeed every bounded continuous periodic function has $E_n(f)\le \Vert f{\Vert }_{\infty }=O(n^{\alpha })$. For an explicit counterexample, on $[-\pi ,\pi ]$ setand extend periodically. This is continuous, but $\omega (h,\delta )\ge 1/\log (e/\delta )$ for $0<\delta <1$, which is not $O({\delta }^{\alpha })$ for any $\alpha >0$. Thus the sign correction is required in the actual printed question, not just in its converted TeX.

For real continuous $2\pi$-periodic $f$, write $E_n$ for best uniform approximation by degree-at-most-$n$ trigonometric polynomials and $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. A quantitative inverse theorem for trigonometric approximation iswhere $C$ is universal. Here is a direct reason for the sum. Choose best approximants $p_j$ and $2^r\le n<2^{r+1}$. WriteThe Bernstein inequality for trigonometric polynomials givesFor $j\ge 2$, monotonicity of best errors bounds $2^{j+1}{E}_{2^{j-1}}$ by $8{\sum }_{\nu =2^{j-2}}^{2^{j-1}-1}{E}_{\nu }$. The first terms are bounded by a fixed multiple of $E_0$. Also $2E_{2^r}\le 4n^{-1}{\sum }_{\nu =0}^n{E}_{\nu }$. Nowproves the theorem, for example with a sufficiently large absolute constant $C=20$.

Suppose, as intended, $E_{\nu }(f)\le A{\nu }^{-\alpha }$ for $\nu \ge 1$, after enlarging $A$ to include any finite exceptional indices, where $0<\alpha <1$. Since ${\sum }_{\nu =1}^n{\nu }^{-\alpha }\le n^{1-\alpha }/(1-\alpha )$, the inverse theorem givesFor $0<\delta \le 1$ take $n=\lfloor 1/\delta \rfloor$. Then $1/(n+1)<\delta \le 1/n$, so monotonicity of the modulus of continuity and $n^{-\alpha }\le 2^{\alpha }{\delta }^{\alpha }$ give $\omega (f,\delta )\le 2^{\alpha }{C}_f{\delta }^{\alpha }$. For $\delta >1$, the separate bound $\omega (f,\delta )\le 2\Vert f{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }{\delta }^{\alpha }$ covers the remaining scales. Therefore the corrected conclusion isNow put $\beta =\log 2/\log 3$. The given best approximant to the Weierstrass function leaves the uniformly convergent tail ${\sum }_{k=m+1}^{\infty }{2}^{-k}\cos (3^{k}x)$. Its supremum norm is exactly $2^{-m}$, attained at $x=0$; henceThe corrected inverse theorem gives $\omega (g,\delta )=O({\delta }^{\beta })$. To show that this order is sharp, set $h_m=\pi /3^{m+1}$. Every term of $g(0)-g(h_m)$ is nonnegative, and for $k\ge m+1$ the cosine equals $-1$ because $3^{k-m-1}$ is odd. ThusGiven small $\delta >0$, select $m$ with $h_m\le \delta <3h_m$. Then $\omega (g,\delta )\ge 4(3\pi {)}^{-\beta }{\delta }^{\beta }$. Combining the estimates gives the precise small-scale orderThe exact approximation error can also be verified independently of the supplied fact: at $x_j=j\pi /3^{m+1}$ the tail has alternating values $(-1{)}^j{2}^{-m}$. There are enough such points in one period for the trigonometric Chebyshev alternation theorem, since $n<3^{m+1}$. This is the positive lacunary trigonometric series mechanism behind the best partial sums.

Take any trigonometric polynomial $t\in {\mathcal{T}}_n$. The reproduction property in (a) givesBy (b),Taking the infimum over $t\in {\mathcal{T}}_n$ yields the polynomial reproduction error bound for this operator. With $E_n(f)={\inf }_{t\in {\mathcal{T}}_n}\Vert f-t{\Vert }_{\infty }$ and $n/m\le M$, it becomesThis is a near-best uniform approximation estimate; it follows from reproduction and boundedness without requiring the de la Vallée Poussin sum itself to be a best approximant.

The Chebyshev alternation theorem says that for $f\in C([a,b])$ a polynomial $q$ of degree at most $n-1$ is a best uniform approximation precisely when there are $n+1$ ordered points $a\le x_0<\cdots <x_n\le b$ such thatThe best polynomial is unique. These equally large alternating extrema are the equal-ripple criterion; the zero-error case is interpreted in the evident way.

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

The bounded-variation space carries the norm $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_{L^1(\Omega )}+|Du|(\Omega )$. Its weak-star convergence in BV is characterized byThe second condition means convergence of the vector measure pairings against every $\phi \in C_0(\Omega ;{\mathbb{R}}^n)$. Equivalently, strong convergence in $L^1$ together with ${\sup }_k|D{u}_k|(\Omega )<\infty$ suffices: integration by parts identifies the limit on smooth compactly supported tests, and uniform approximation extends this to $C_0$ tests.

The bounded-variation compactness theorem says thatguarantees a subsequence convergent in weak-star convergence in BV on a bounded Lipschitz domain. This is the uniform criterion for relative sequential compactness. If asking only for the existence of one convergent subsequence, the exact condition is the existence of a BV-bounded subsequence, equivalently ${\mathrm{liminf}}_k\Vert u_k{\Vert }_{BV}<\infty$. The entire sequence need not be bounded: alternating zero functions and constants tending to infinity gives a simple example. Conversely, a convergent subsequence has bounded $L^1$ norm and bounded total variation seminorm by the Uniform boundedness principle for its derivative-measure pairings.

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 $n$.

A linear map $U:C(K,\mathbb{R})\to C(K,\mathbb{R})$ is positive when $f\ge 0$ implies $Uf\ge 0$. This is positivity for the pointwise order, not positivity of a quadratic form. On a compact space, $|Uf|\le U|f|\le \Vert f{\Vert }_{\infty }U1$, so $\Vert U\Vert =\Vert U1{\Vert }_{\infty }$. This makes positive maps especially useful for uniform approximation.