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For $f\in C([-1,1])$, the equation $((1-x^2)y^{\prime }{)}^{\prime }=f$ has a bounded solution on $(-1,1)$ if and only if ${\int }_{-1}^{1}f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y^{\prime }$ has finite endpoint limits, and any nonzero endpoint limit would force logarithmic growth in $y$. Both limits must therefore vanish, yielding the zero-mean condition. For sufficiency set $F(x)={\int }_{-1}^{x}f(t)dt$ and $y(x)=C+{\int }_0^{x}F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\le \Vert f{\Vert }_{\infty }\min (1+s,1-s)$, so $y^{\prime }$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\frac{1}{2}\log ((1+x)/(1-x))$, proving uniqueness up to a constant. This is the Fredholm alternative for the zero eigenvalue of this singular Sturm-Liouville problem.

The eigenfunction expansion in Legendre polynomials is $y=C-{\sum }_{k\ge 1}{P}_k{\int }_{-1}^{1}f{P}_k/[k(k+1){\int }_{-1}^1{P}_k^2]$. It follows by projecting onto the eigenfunctions with eigenvalues $-k(k+1)$. Completeness follows because the Legendre polynomials span the polynomials, which are dense by the Weierstrass approximation theorem.

Continuity only in the open interval requires more care. Even if $f$ is integrable and has zero mean, the primitive of $F/(1-x^2)$ must still be bounded at both ends. For example, $F(x)=1/\log (e/(1-x^2))$ has zero endpoint limits and absolutely integrable derivative $f=F^{\prime }$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.

On a compact interval, positive quadrature rule weights whose sum is the interval length define uniformly bounded functionals on continuous functions. If the rules eventually integrate every fixed polynomial exactly, the Weierstrass approximation theorem proves convergence for every continuous function.

If each $(n+1)$-node quadrature rule is exact through degree $n$ and has positive weights, then it converges on every continuous function. The Weierstrass approximation theorem reduces the error to the uniform polynomial-approximation error, while exactness on constants controls the sum of the weights.

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 $C=1$.

For a real continuous function on a compact interval, choose its best uniform approximation $p_n^{\ast }$ of polynomial degree at most $n$. If its error is nonzero, the Chebyshev alternation theorem supplies $n+2$ ordered alternating extrema. The intermediate value theorem gives a zero of $f-p_n^{\ast }$ in each intervening interval. At these $n+1$ distinct points, the polynomial interpolant is exactly $p_n^{\ast }$. If the best error is zero, arbitrary distinct nodes suffice. The Weierstrass approximation theorem shows that these function-dependent interpolants converge uniformly. The nodes depend on the target function; the conclusion does not provide one universal node sequence for every continuous target.

Two probability measures supported on one compact interval are equal if all their moments are equal. Indeed, equality on monomials gives equality on polynomials, the Weierstrass approximation theorem extends it to every continuous function on the interval, and compactly supported continuous functions determine a finite Borel measure.

The orthogonal projection is $p_n={\sum }_{r=0}^n(u,{\varphi }_r){\varphi }_r$. Subtract it from $u$ and use the orthonormal basis relations to obtainThis is the requested Bessel inequality.

For the infinite tail, use completeness of the orthonormal polynomial family in the usual weighted L2 space. The relevant density facts are that continuous functions are dense for the finite measure $w(x)dx$, and the Weierstrass approximation theorem makes polynomials uniformly dense in continuous functions on the compact interval. Since $\int w<\infty$, uniform approximation also gives weighted L2 norm approximation. Parseval's identity therefore givesHere finiteness of the weight integral is also necessary for the constant polynomial to belong to the stated Hilbert space. Without completeness of the polynomial family, an additional squared component orthogonal to every polynomial would remain; Bessel inequality alone does not justify equality.

The even-power family is dense on $[1,2]$. It is the algebra $A=\{p(t^2):p\in \mathbb{C}[s]\}$. It contains constants, is closed under complex conjugation, and separates points because the real function $t\mapsto t^2$ is injective on this positive interval. The complex Stone-Weierstrass theorem therefore applies.

Equivalently, the squaring map is a homeomorphism from $[1,2]$ onto $[1,4]$. Given $f\in C([1,2])$, approximate the continuous function $s\mapsto f(\sqrt{s})$ uniformly on $[1,4]$ by complex polynomials, using the Weierstrass approximation theorem on the real and imaginary parts. Substitution $s=t^2$ produces exactly the required even-power approximants. The positive interval is what lets even powers distinguish all its points.

First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y^{\prime }$ has finite endpoint limits. A nonzero limit would make $y^{\prime }$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequentlyThis is orthogonality to the constant eigenfunction $P_0$, hence the Fredholm alternative condition. It is also sufficient under the stated endpoint regularity: putIf $M=\max |f|$, the zero-mean condition gives $|F(s)|\le M\min (1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z^{\prime }{)}^{\prime }=0$, whose other solution is $\frac{1}{2}\log ((1+x)/(1-x))$.

For the requested eigenfunction expansion, setThe Legendre polynomials span all polynomials, so their completeness follows from the Weierstrass approximation theorem. Projecting $\mathcal{L}y=f$ onto each eigenfunction gives $-k(k+1)y_k=f_k$ for $k\ge 1$. The zero mode is free, while $f_0=0$. Thus the full family isThe formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\le 1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the Cauchy-Schwarz inequality, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second differentiation.

There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, letThis $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm 1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y^{\prime }$ haswhose integral diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makesbounded as $x$ approaches both endpoints. The Bounded solvability for the Legendre operator criterion above covers the intended regular forcing and makes this endpoint limitation explicit.

Composition with $t\mapsto t^{1+{\alpha }_n}$ preserves continuity, so $A_n$ maps the space of continuous functions on a compact space $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+{\alpha }_n}$ maps $[0,1]$ onto itself,For $0<t\le 1$, the mean value theorem applied to $r\mapsto t^{1+r}$ givesThe same bound holds at $t=0$. Each $x\in C([0,1])$ has uniform continuity by the Heine-Cantor theorem. With its modulus of continuity ${\omega }_x(\delta )={\sup }_{|s-t|\le \delta }|x(s)-x(t)|$, we obtainAlternatively, the bound proves convergence on every polynomial; the Weierstrass approximation theorem makes these a dense subset of $C([0,1])$, and the Banach-Steinhaus convergence theorem with $\Vert A_n\Vert =1$ gives the same conclusion.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

Choose a best uniform approximation $p_n\in {\mathcal{P}}_n$. It exists because this vector subspace is a finite-dimensional vector space and a minimizing sequence $q_j$ is bounded: $\Vert q_j{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }+\Vert f-q_j{\Vert }_{\infty }$. Finite-dimensional equivalence of norms then supplies a convergent subsequence of coefficients. If $E_n(f)>0$, the Chebyshev alternation theorem gives $n+2$ alternating error extrema. The intermediate value theorem supplies a zero $x_{n,k}$ of $f-p_n$ in each of the $n+1$ intervening open intervals. These interpolation nodes are distinct, and uniqueness of polynomial interpolation givesThe convergence follows from the Weierstrass approximation theorem. If $E_n(f)=0$, choose any $n+1$ distinct nodes: the polynomial interpolant is still $f=p_n$. This constructs interpolation nodes from best uniform approximation; the nodes depend on $f$, and no universal node system for all continuous functions is asserted.

Write $W={\int }_{-1}^{1}w(x)dx$, finite and positive. Exactness on the constant polynomial gives ${\sum }_i{a}_i^{(n)}=W$. Positivity therefore implies the uniform operator bound $|I_n(h)|\le W\Vert h{\Vert }_{\infty }$, and the integral has the same bound.

For a continuous $f$ and $ϵ>0$, the Weierstrass approximation theorem gives a fixed polynomial $Q$ of degree $m$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For every $n\ge m$, polynomial exactness gives $I_n(Q)=I(Q)$. ThusSince $ϵ$ is arbitrary, $\boxed{I_n(f)\to I(f)}$ for every continuous $f$. This is convergence of positive quadrature on continuous functions. The argument requires nodes in $[-1,1]$, implicit in a rule defined on all continuous functions there; otherwise their function values would not even be specified by the stated domain. Exactness alone would not provide the norm bound if arbitrarily signed weights were allowed.

For each $n$, choose the best uniform approximation $p_n^{\ast }$. If its error is nonzero, take its $n+2$ alternating extrema. The intermediate value theorem supplies a point $y_j$ strictly between the $j$th and $(j+1)$st extrema such that $f(y_j)=p_n^{\ast }(y_j)$, for $0\le j\le n$. These $n+1$ points are distinct. The polynomial interpolation problem at them has a unique solution of degree at most $n$, so its solution is exactly $p_n^{\ast }$. This constructs the interpolation nodes from best uniform approximation.

If $E_n(f)=0$, then $f$ itself belongs to the polynomial space, and any $n+1$ distinct nodes give the same conclusion. In either case, take $x_{n,j}=y_j$ or the arbitrary nodes just described. ThenThe Weierstrass approximation theorem gives $E_n(f)\to 0$; alternatively, the bound derived in part 4(b), combined with uniform continuity, gives this convergence directly. ThereforeThe nodes may depend on $f$ and on $n$. The conclusion does not assert a universal sequence of interpolation grids working for every continuous target.

Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.

Use the unscaled Finite Fresnel integrals of this paper. Termwise integration of the absolutely convergent Taylor series for sine and cosine gives the full small-argument expansionsIn particular,These Taylor series converge at every finite argument, although cancellation makes them unattractive for large arguments.

For the large-argument asymptotic expansions, combine the two Finite Fresnel integrals into $C+iS$. The standard Fresnel integral givesFor the tail $J(\lambda )={\int }_{\lambda }^{\infty }{e}^{i{t}^2}dt$, integration by parts using $(e^{i{t}^2}{)}^{\prime }=2it{e}^{i{t}^2}$ givesFor example $a_1=1/4$, $a_2=-3i/8$, and $a_3=-15/16$. Repeated integration by parts bounds the remainder after any fixed number of terms by the order of the next term. Equivalently the Fresnel tail expansion isHere $(-1)!!=1$. The first three terms of the requested cosine Fresnel integral, counting its constant term, areFor comparison, retaining one further correction in both Finite Fresnel integrals givesUnlike the small-argument Taylor series, these large-argument asymptotic expansions must be truncated; adding indefinitely many terms at a fixed argument does not improve the approximation.

A calculator needs elementary arithmetic and its built-in sine and cosine, together with stored approximation coefficients. A convenient global construction uses compactified approximation from endpoint asymptotics. Define the exact real tail functions byThen $P(0)=Q(0)=A$, while $P\sim 1/(2\lambda )$ and $Q\sim 1/(4{\lambda }^3)$ at infinity. Consequentlyextend continuously to the compact interval $0\le u\le 1$, with $p(1)=1/2$ and $q(1)=1/4$. Choose stored polynomial approximations $p_N,q_N$ on that interval, constrained by the small-argument Taylor series and large-argument asymptotic expansions, and calibrated across the intervening interval. The resulting elementary formula isUniform errors ${\delta }_p,{\delta }_q$ in the stored polynomial approximations give $|C_{\mathrm{calc}}-C|\le {\delta }_p+{\delta }_q$ for every positive argument. Existence of arbitrarily accurate such approximations follows from the Weierstrass approximation theorem. A Padé approximant with a denominator having no positive zero is an alternative implementation. Matching the two end expansions alone is not an error guarantee: the middle interval must also be controlled. For very small arguments, switching to the short Taylor series avoids cancellation. Ordinary finite-precision limitations in evaluating ${\lambda }^2$ and reducing its trigonometric phase remain those of the calculator itself.

For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

The Weierstrass approximation theorem states that the polynomials are a dense subspace of $C[0,1]$ in the supremum norm. By linearity and (c), $B_{n}p\to p$ uniformly for every polynomial $p$. Given $f\in C[0,1]$ and $\epsilon >0$, choose a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. The operator norm estimate in (a) givesFirst let $n\to \infty$, then $\epsilon \downarrow 0$. Hence $\Vert B_{n}f-f{\Vert }_{\infty }\to 0$ for every continuous $f$. This is an instance of extension of approximation operators from a dense subspace.

The Weierstrass approximation theorem says that for every real continuous function $f$ on $[0,1]$ and every $\epsilon >0$, there is a real polynomial $P$ with ${\sup }_{[0,1]}|f-P|<\epsilon$. Equivalently, polynomials are dense in $C[0,1]$ in the uniform norm.

This also forces $f=0$. Indeed, $h(t)=f(\sqrt{t})$ is a continuous function on $[0,1]$. Approximate $h$ uniformly by polynomials $P_k(t)$ using the Weierstrass approximation theorem. Then $P_k(x^2)\to f(x)$ uniformly, and the even moments give $\int f(x)P_k(x^2)dx=0$. The estimate from part (i), with $P_k(x^2)$ in place of $P_k(x)$, proves $\int f^2=0$ and hence the conclusion.