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The Korovkin theorem says that a sequence of positive linear operators on continuous functions on $[0,1]$ converges uniformly to the identity on every continuous function if it does so on $1$, $x$ and $x^2$. Necessity is immediate; the quadratic barrier argument establishing sufficiency is given in Solution 6.

There is an endpoint misprint in the original PDF: the repeated knots at the right end must equal $1$, not $0$. Otherwise the advertised nondecreasing spline knot sequence on $[0,1]$ does not exist. Use the corrected clamped spline knot sequences, with fixed order $k$ and maximum gap $h_n\to 0$. Let $m_n$ denote the number of B-splines for the $n$th sequence. Nonnegativity and partition of unity show that the Schoenberg spline operator $V_n$ is positive and $V_{n}1=1$.

The monomial B-spline coefficients reproduce the first two nonconstant test polynomials. PutThe pair sum uses $r<s$, as in the PDF; the converted TeX incorrectly includes $r=s$. Every sampling point ${\tau }_j$ and every knot entering these coefficients lie in $[t_j,t_{j+k}]$, whose length is at most $k{h}_n$. Thus $|{\tau }_j-a_{1,j}|\le k{h}_n$. Also all these numbers belong to $[0,1]$, soThe positive B-spline weights giveTogether with $V_{n}1=1$, the three limits proveUnder the usual interior knot multiplicities at most $k-1$, the splines are continuous and the quoted form of the Korovkin theorem applies directly. If full interior multiplicity is allowed, the output can be discontinuous. The same positive-operator quadratic barrier proof still works with bounded output functions and the supremum norm, and proves exactly the stated uniform error limit without asserting continuity of each output. Indeed the stronger local-support error bound for a Schoenberg spline operator, $\Vert V_{n}g-g{\Vert }_{\infty }\le \omega (g,k{h}_n)$, follows by comparing $g(x)$ with each active sample value. The restriction $k\ge 3$ is needed for the displayed quadratic reproduction argument, not for this direct support estimate.

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.

Interpret convergence in $C[0,1]$ as uniform convergence, and work first with real continuous functions. Set $e_{n,j}=U_n(p_j)-p_j$, for $j=0,1,2$, and put $M=\Vert f{\Vert }_{\infty }$. The given barriers have one $\gamma \ge 0$ independent of both $t$ and $n$. A positive linear operator on continuous functions preserves non-strict inequalities, so application of $U_n$ and evaluation at $t$ giveStrict pointwise inequalities need not remain strict under a positive operator; non-strict ones suffice. Linearity and the cancellation of the exact moments yieldSubtracting $f(t)$ rather than $f(t)U_{n}1(t)$ therefore gives the quadratic barrier estimate for positive approximation operators:For fixed $\epsilon$, all three moment errors tend uniformly to zero. Thus ${\mathrm{limsup}}_n\Vert U_{n}f-f{\Vert }_{\infty }\le \epsilon$. Since $\epsilon$ was arbitrary,This completes the Korovkin theorem proof. Complex-valued functions follow by applying the real result to real and imaginary parts. Uniform convergence of the three test functions is essential to this uniform conclusion; purely pointwise test convergence should not be silently substituted.

The Korovkin theorem states that if positive linear operators on continuous functions $L_n:C[0,1]\to C[0,1]$ satisfythen $\Vert L_{n}f-f{\Vert }_{\infty }\to 0$ for every real continuous function $f$.

Let $r=k-1\ge 2$ and $h_n=|{\Delta }_n|$. For the B-splines in this question, $N_{i,n}\ge 0$ and their sum is one on the basic interval. Thus the Schoenberg spline operator $V_n$ is linear and positive. To verify the three test functions, suppress the index $n$ and write the Marsden identity with ${\omega }_i(x)={\prod }_{j=1}^r(x-t_{i+j})$. Comparing the coefficients of $x^r,x^{r-1},x^{r-2}$ gives the monomial B-spline coefficientswhereThe first coefficient is the partition of unity; the second is the Greville abscissa. Every internal knot and the sample point ${\tau }_i$ lie in $[t_i,t_{i+k}]$, of width at most $k{h}_n$. Consequently $|{\tau }_i-{\xi }_i|\le k{h}_n$. Since all these numbers lie in $[0,1]$, for any two internal knots $u,v$ we also haveAveraging gives $|{\tau }_i^2-q_i|\le 2k{h}_n$. Nonnegative weights of sum one now yieldHere the spline order $k$ is fixed, so all three errors tend to zero. The Korovkin theorem therefore provesNo special selection such as the Greville abscissae is required for the sample points. In fact the local-support error bound for a Schoenberg spline operator also gives $\Vert V_{n}f-f{\Vert }_{\infty }\le \omega (f,k{h}_n)$. Under the usual continuous-spline convention the operators have the codomain required by the stated theorem. If the permissive knot notation is taken to allow full interior multiplicities and hence discontinuous splines, the same conclusion in the uniform norm still follows from this direct bound, although each approximant need not then be continuous.

Apply the periodic Korovkin theorem to the constant sequence $U_n=U$. Since it fixes the three test functions, the theorem gives $Uf=f$ for every $f$.

One can also see the rigidity directly. For fixed $x$, the positive linear functional $L_x(f)=(Uf)(x)$ has $L_x(1)=1$ and is represented by a probability measure. The fixed sine and cosine imply $L_x(1-\cos (t-x))=0$. The integrand is nonnegative with unique zero $x$ on the circle, so the measure is ${\delta }_x$. Thus $(Uf)(x)=f(x)$ at every point. The operator is uniquely the identity: $\boxed{U=I}$.

Averaging the given Dirichlet kernels gives the normalized Fejér kernelThe geometric-series identity, or summing the sine terms, proves the second equality. At multiples of $2\pi$ use its removable value $n/(2\pi )$. Therefore $K_n\ge 0$, andis a positive linear operator on continuous functions. The Fourier partial sums, viewed as orthogonal projections, fix $1$, so ${\sigma }_{n}1=1$ and $\int K_n=1$. They also fix $\sin x$ and $\cos x$ for degrees at least one, whereas $s_0$ annihilates them. ConsequentlyAll three periodic Korovkin theorem tests converge uniformly. Hence Fejér sums converge uniformly for every continuous periodic function:

For positive linear operators on continuous functions, a general Korovkin theorem, in its compact Korovkin test space form, can be stated using a test space $H\subset C(K,\mathbb{R})$, where $K$ is a compact Hausdorff space. Assume $H$ contains a strictly positive function $h_0$. Require that, for every $x\in K$, the only positive linear functional $L$ on $C(K)$ with $L(h)=h(x)$ for all $h\in H$ is evaluation at $x$. Then convergence on the test space implies convergence on every continuous function:By the Riesz-Markov-Kakutani representation theorem, the functional condition equivalently says that a positive measure with these test moments must be ${\delta }_x$. This permits arbitrary test families, not just the interval tests $1,x,x^2$.

For completeness, let $c={\min }_K{h}_0>0$. Positivity gives $U_{n}1\le c^{-1}{U}_n{h}_0$, so the operator norms are uniformly bounded. If the conclusion failed for some $f$, choose $x_n$ where its error is bounded away from zero. The evaluation functionals $L_n(g)=U_{n}g(x_n)$ have uniformly bounded norm. Compactness of $K$ and the Banach-Alaoglu theorem give a subnet on which $x_n\to x$ and $L_n$ converges in the weak-star topology to a positive $L$. For every $h\in H$, uniform test convergence gives $L(h)=h(x)$. The hypothesis forces $L(f)=f(x)$, contradicting the chosen errors. Complex-valued functions follow by applying the real result to their real and imaginary parts.

A useful concrete sufficient condition is that $1\in H$ and for each $x$ there is $g_x\in H$ with $g_x\ge 0$ and zero set exactly $\{x\}$. Its zero moment forces the representing measure to be supported at $x$, while the constant moment fixes its mass.

For the circle, take $H=\mathrm{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos (t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus the periodic Korovkin test set isUniform convergence on these three functions is sufficient for uniform convergence on $C(\mathbb{T})$.

The Korovkin theorem states that positive linear operators $L_n:C[0,1]\to C[0,1]$ converge uniformly to the identity on every continuous function if they do so on the three test functions $1,t,t^2$.

The Schoenberg spline operator is positive and $V_{n}1=1$. If ${\xi }_i=a_{1,i}$, then both ${\tau }_i$ and ${\xi }_i$ lie in $[t_i,t_{i+k}]$, soBecause $a_{2,i}$ averages products of knots in the same interval and all points lie in $[0,1]$,The Korovkin theorem now provesfor every $f\in C[0,1]$.

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.