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Fix $\epsilon >0$ and put $\eta =\epsilon /2$. For each pair $s,t\in T$, choose $a_{s,t}\in A$ whose errors at these two points are smaller than $\eta$. By continuity, there is a neighborhood $U_{s,t}$ of $t$ on which $a_{s,t}>f-\eta$. For a fixed $s$, these neighborhoods cover the compact space $T$. Choose a finite subcover $U_{s,t_1},\dots ,U_{s,t_m}$ and formThen $b_s>f-\eta$ everywhere, while $b_s(s)<f(s)+\eta$, because every chosen function has error smaller than $\eta$ at $s$. Since $b_s$ is continuous, there is a neighborhood $W_s$ of $s$ on which $b_s<f+\eta$.

A second use of compactness supplies $W_{s_1},\dots ,W_{s_q}$ covering $T$. SetEvery $b_{s_i}$ is above $f-\eta$, so their minimum is also above it. At each point at least one $b_{s_i}$ is below $f+\eta$, and hence so is their minimum. This proves $\Vert a-f{\Vert }_{\infty }\le \eta <\epsilon$. The two finite operations are the essential mechanism in lattice approximation from two-point approximation; no closure under addition or multiplication has been assumed.

For the corollary take $A$ to consist of all continuous piecewise linear functions on $[0,1]$. On a common finite partition, the maximum or minimum of two affine pieces can change which piece it selects only at their crossing. Adding those finitely many crossings to the partition shows that $A$ is closed under both operations. For two distinct points, the affine polynomial through the two prescribed values belongs to $A$ and matches $f$ exactly there; for equal points use a constant function. The proved criterion therefore gives density of continuous piecewise linear functions in $C[0,1]$ in the supremum norm. Concretely, interpolation on a partition of maximum gap $h$ also gives error at most $\omega (f,h)$ by uniform continuity.

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.

Fix a clamped nondecreasing spline knot sequence $t_1,\dots ,t_{n+k}$ on $[a,b]$, with $t_1=\cdots =t_k=a$ and $t_{n+1}=\cdots =t_{n+k}=b$. For continuous splines assume interior multiplicities at most $k-1$, and assume $t_i<t_{i+k}$. An order-$k$ spline has degree at most $k-1$ on each nonempty knot interval; at a knot of multiplicity $r$ it has $k-1-r$ continuous derivatives. Its dimension is $n$, and a normalized B-spline basis $N_1,\dots ,N_n$ has local support, nonnegative values and sum one on $[a,b]$.

These properties are visible in the Cox-de Boor recurrence. Starting with interval indicators at order one, setwith a term of zero denominator defined to be zero. On each contributing support the weights are nonnegative, and summing the recurrence makes the coefficients of each lower-order B-spline add to one. This gives partition of unity and $\mathrm{supp}{N}_{i,k}=[t_i,t_{i+k}]$, with endpoint values defined by limits.

At distinct ordered sites $x_1<\cdots <x_n$, interpolation is the linear systemIt has a unique solution for every data vector exactly when the B-spline collocation matrix is invertible. The Schoenberg–Whitney theorem characterizes this by $N_i(x_i)>0$ for every $i$; away from clamped endpoints this is $t_i<x_i<t_{i+k}$. At the clamped endpoints the first and last B-splines have value one, so the endpoint version uses this positive-value formulation rather than the strict inequalities. Both ordering and the support conditions matter. A spline space with interior knots is generally not a Chebyshev system on the entire interval: a locally supported B-spline vanishes on an interval. Thus arbitrary distinct nodes need not permit interpolation. Merely specifying all data sites as knots also need not fix the remaining freedom; for example, cubic interpolation with simple interior knots at $m-2$ data sites has dimension $m+2$, and two extra boundary conditions are needed to interpolate $m$ values uniquely.

The positivity governing this existence theorem is total nonnegativity of B-spline collocation matrices: every square minor with increasing row and column indices is nonnegative. In spline terminology this is often called total positivity, but it permits zero minors caused by disjoint supports. It is not the stronger assertion that every minor is strictly positive. A useful proof uses knot insertion. Each refinement step changes coefficients bywith the unchanged endpoint pieces. The refinement matrix is nonnegative bidiagonal, and its minors are nonnegative: any nonzero determinant has its allowed diagonal matching and is a product of nonnegative entries. Products preserve this property by the Cauchy–Binet formula. Insert each sampling site to full multiplicity, so evaluation at that site becomes selection of an ordered refined coefficient. The collocation matrix is then an ordered row submatrix of the refinement product, proving total nonnegativity. The strict support criterion is the strict part of this argument: an ordered path through the consecutive support intervals exists exactly when every diagonal B-spline value is positive, giving the Schoenberg–Whitney theorem determinant criterion.

If $A$ is invertible, its determinant is positive and the adjugate identity yieldsThis checkerboard inverse of a totally nonnegative matrix is the central stability fact. With alternating data $e_i=(-1{)}^i$, the absolute row sums satisfyFor general data, the cardinal splines are ${\ell }_i(x)={\sum }_j{N}_j(x)(A^{-1}{)}_{ji}$. The spline interpolation operator and its exact Lebesgue constant of interpolation areThe equality for the operator norm follows by extending the signs at distinct sites to a continuous unit-norm data function, as in Solution 2. That solution also shows the last inequality can be very strict. Stability concerns the range function as well as its coefficients.

An optimal interpolation set must be specified relative to an objective. For the normalized B-spline coefficients, every admissible sampling set satisfiessince $c(s)=A_{\mathbf{x}}^{-1}(s(x_i))$ and sampling cannot increase the supremum norm. If a unit-norm spline $s_{\ast }$ has $n$ ordered alternating extrema $s_{\ast }(x_i)=(-1{)}^i$ at admissible sites, the checkerboard identity givesTaking the largest row shows $\Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }\le \kappa (\mathcal{S})$, so equality holds. This proves the optimal spline coefficient interpolation sites property and the Chebyshev spline coefficient and dual norm equality. In particular, for the cubic space of Solution 1 the alternating extrema $-1,-2/3,0,2/3,1$ satisfy the support conditions. Its coefficients have maximum magnitude $11/2$, so these sites are optimal for coefficient recovery andThe argument proves optimality whenever this admissible equioscillating spline is available; it does not identify extrema of an arbitrary spline as optimal sites, or transfer coefficient optimality automatically to the Lebesgue constant of interpolation.

For the intrinsic interpolation objective, minimize ${\Lambda }_{\mathbf{x}}$ itself. There is a minimizer for a fixed finite-dimensional continuous spline space. Indeed the sites lie in a compact ordered simplex. On admissible sets, inverse entries and hence ${\Lambda }_{\mathbf{x}}$ vary continuously. As a collocation matrix tends to a singular one, its inverse norm diverges. The uniform-norm stability of a B-spline basis gives the complementary boundchoose sample signs attaining a largest inverse row, extend them to a continuous unit-norm function, and use $\Vert Tc{\Vert }_{\infty }\ge \Vert c{\Vert }_{\infty }/\kappa (\mathcal{S})$. Thus a minimizing sequence with bounded Lebesgue constants of interpolation cannot approach a singular site set. A convergent subsequence has an admissible limit attaining the minimum.

Fekete interpolation sites for a continuous function space offer another useful, constructive criterion: maximize the absolute evaluation determinant. A nonzero maximum exists by compactness and linear independence. Replacing one row by evaluation at $x$ multiplies the determinant by the corresponding cardinal function, so maximality gives $|{\ell }_i(x)|\le 1$, and therefore ${\Lambda }_{\mathbf{x}}\le n$. These sites are determinant-optimal, with a proven interpolation bound; one should not claim without further argument that they minimize ${\Lambda }_{\mathbf{x}}$. Greville abscissae provide inexpensive admissible sites for the usual continuous clamped spline spaces of order at least two. Admissibility alone does not give the best stability bound.

Finally, the coefficient condition number of a normalized B-spline basis is bounded by a constant depending only on order. To see the mechanism, for each coefficient choose a longest nonempty knot cell in that B-spline's support. If its length is $h$ and the support width is $W$, then $W/h\le k$. On that cell the spline is a degree-at-most-$k-1$ polynomial; repeated Markov polynomial inequalities bound its $r$th derivative by $C_k{h}^{-r}\Vert s{\Vert }_{\infty }$. In the De Boor–Fix spline coefficient functional, the accompanying derivative of the knot polynomial is bounded by $C_k{W}^r$. Each term is consequently at most $C_k(W/h{)}^r\Vert s{\Vert }_{\infty }$, yielding $\Vert c(s){\Vert }_{\infty }\le C_k\Vert s{\Vert }_{\infty }$ independently of the number and spacing of knots. This is basis stability; it does not bound $\Vert A_{\mathbf{x}}^{-1}\Vert$ at arbitrarily poor sampling sites.

For approximation, reproduction gives the Lebesgue inequality for approximation projectorsAs the maximum knot gap tends to zero with fixed order, spline quasi-interpolation gives an approximant with error at most $\omega (f,kh)$, so the infimum tends to zero. A sequence of uniformly bounded interpolation linear operators therefore converges to every continuous target. Local support and a stable choice of sites make spline interpolation effective; mesh refinement without control of the interpolation operator norm is not by itself a convergence proof.

Write $s=A(S)=4\pi$, $L=L(\partial U)$, and $a=\min \{A(U),A(S\setminus U)\}$. We first prove a constant-version spherical isoperimetric inequality directly, including regions with several boundary components.

If $L\ge 1$, the arithmetic bound $A(U)A(S\setminus U)\le s^2/4=4{\pi }^2$ already gives $A(U)A(S\setminus U)\le 4{\pi }^2{L}^2$. If $L=0$, a region with smooth boundary is empty or the whole connected sphere up to boundary-null changes, and its area product is zero. For $0<L<1$, write the smooth boundary as finitely many disjoint simple closed curves ${\gamma }_j$ of lengths $l_j$. Fix a point $p_j$ on each curve. The shorter boundary arc from $p_j$ to any other point has length at most $l_j/2$, so ${\gamma }_j$ is contained in the closed geodesic ball of radius $l_j/2$ about $p_j$.

The complement of that small ball is connected and lies entirely on one side of ${\gamma }_j$. The Jordan curve theorem therefore gives a disk $E_j$ bounded by ${\gamma }_j$ contained in the ball. Its area satisfiesThe indicator function of $U$ and the sum ${\sum }_j{1}_{E_j}$ have the same jumps modulo two across every boundary curve. Their difference modulo two is consequently constant throughout the sphere off the boundary. Thus either $U$ or its complement lies in ${\bigcup }_j{E}_j$, up to its area-zero boundary. This also handles nested curves and disconnected regions. It follows thatCombining the cases, one valid universal choice is $\kappa =4{\pi }^2$. Sharpness is not required.

For the inclusion estimate put $x=A(V)$, $d=A(D)>0$ and $y=A(V\cap D)$. Splitting $D$ into its portions inside and outside $V$ givesBecause $0\le A(D\setminus V)\le s-x$ and $0\le A(D\cap V)\le x$, this expression lies between $-x(s-x)$ and $x(s-x)$. Apply the preceding spherical isoperimetric inequality to $V$ and divide by $sd$:Thus the constant in the spherical inclusion area discrepancy is independent of the nonempty open set $D$.

Let $g$ and $\tilde{g}$ be the old and new smooth Riemannian metrics. Their area forms satisfyThe determinant ratio is invariant under changes of coordinates, so $\theta$ is a globally defined smooth function. The layer cake representation follows from $\theta (x)={\int }_0^{\infty }{1}_{\{\theta (x)>t\}}dt$ and Tonelli's theorem:Smoothness makes every $D_t$ open. Compactness of the sphere gives $0<m=\min \theta \le M=\max \theta <\infty$. Multiply the inclusion bound by $A(D_t)$ before integrating; for empty $D_t$ the resulting bound is trivially zero. For $t<m$, $D_t=S$ and the discrepancy is exactly zero; for $t>M$, $D_t$ is empty. ThereforeDividing by $\tilde{A}(S)>0$ provesThis constant depends only on the two metrics. If $\theta$ is constant the normalized areas agree exactly, as the bound also shows.

Use the reduced-denominator function unbounded on every interval:Every value is a finite real number. In any nondegenerate interval $[a,b]\subseteq [0,1]$, its interior contains infinitely many rational numbers, whereas there are only finitely many reduced fractions in $[0,1]$ with denominator at most any fixed $M$. Some rational number in $(a,b)$ therefore has denominator exceeding $M$. Hence the function is unbounded on every such interval. It reverses the positive values of the Thomae function rather than requiring an infinite value at any point.

The mean value theorem states that, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, there is $c\in (a,b)$ such thatTo deduce it from Rolle's theorem, subtract the secant line: $k(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a)$. This function has the same continuity and differentiability properties and satisfies $k(a)=k(b)=0$. Rolle's theorem gives an interior point with $k^{\prime }(c)=0$, which is precisely the required equality.

If $h^{\prime }=0$ on $\mathbb{R}$, apply the mean value theorem on any $[x,y]$ with $x<y$. It gives $h(y)-h(x)=(y-x)h^{\prime }(c)=0$. Hence $h$ is constant.

For the differential equation, the product rule givesThe preceding result makes this product constant. Thus all solutions are $f(x)=C{e}^{ax}$, and differentiation verifies every such solution.

For the integral assertion, continuity ensures that $f$ is Riemann integrable on every finite closed interval. Use the usual orientation for an integral with negative upper endpoint. For $h\ne 0$, additivity of the integral givesThe absolute value of this expression is at most the supremum of $|f(t)-f(x)|$ on the segment joining $x$ and $x+h$. Continuity at $x$ makes this bound tend to zero for either sign of $h$. Consequently $F^{\prime }(x)=f(x)$, proving this form of the fundamental theorem of calculus.

Finally, a differentiable solution of the integral equation is continuous, so the fundamental theorem of calculus yields $g^{\prime }=g$; setting $x=0$ gives $g(0)=A$. The differential-equation calculation above now forcesConversely, $A+{\int }_0^{x}A{e}^{t}dt=A+A(e^x-1)=A{e}^x$, so this function satisfies the original integral equation. It is unique because every differentiable solution has already been forced to this form and this initial condition.

For a function with continuous derivatives through order $m+1$ on the interval between $0$ and $x$, the Taylor formula with integral remainder isThis is Taylor's theorem about zero; translation gives the formula about any other point. To prove it, the case $m=0$ is the fundamental theorem of calculus. For $m\ge 1$, integration by parts givesSubstitution into the formula at order $m-1$ proves the formula at order $m$ by induction. The integral estimate, valid also for negative $x$, isFor the smooth function satisfying $f^{(3)}=f$, each fixed derivative is continuous on $[-R,R]$ and hence is bounded there by the extreme value theorem. Thus a bound $M_j$ exists for every $j$. Differentiating the differential equation repeatedly gives $f^{(j+3)}=f^{(j)}$. Only the three functions $f,f^{\prime },f^{\prime \prime }$ are needed: take $M$ to be the maximum of their absolute-value bounds on $[-R,R]$. Then $|f^{(j)}(x)|\le M$ simultaneously for every $j\ge 0$ and $|x|\le R$.

At zero the derivatives consequently repeat the values $1,0,0$. Taylor's theorem therefore givesThe last limit of a sequence follows because the ratio of consecutive majorants is $R/(m+2)$, eventually less than $1/2$. Since $R$ is arbitrary, Taylor expansion from a periodic derivative equation provesThis proves convergence to the function using a remainder bound; it does not assume beforehand that a smooth function equals its Taylor series.

If $f:A\to B$ is a monomorphism, each function $\mathcal{C}(U,A)\to \mathcal{C}(U,B)$ given by postcomposition with $f$ is injective. Consequently, if two natural transformations into $H_A$ become equal after $H_{\bullet }(f)$, their components are equal, and the transformations themselves are equal. This proves that $H_{\bullet }(f)$ is a monomorphism in the presheaf category.

Conversely, suppose $H_{\bullet }(f)$ is a monomorphism and $u,v:U\to A$ satisfy $fu=fv$. Then $H_{\bullet }(f)H_{\bullet }(u)=H_{\bullet }(f)H_{\bullet }(v)$. Cancellation gives $H_{\bullet }(u)=H_{\bullet }(v)$, and faithfulness of the Yoneda embedding gives $u=v$. Hence $H_{\bullet }(f)$ is monic exactly when $f$ is monic.

For a presheaf of sets on a topological space $P$, define $LP=P(\varnothing )$ and $RP=P(S)$. Then the adjoints to constant presheaves on open sets areTo prove the first adjunction, a natural transformation $a:P\to \Delta A$ must satisfy $a_U=a_{\varnothing }\circ {\mathrm{res}}_{U,\varnothing }$ by naturality at the inclusion $\varnothing \subseteq U$. Hence it is uniquely determined by a function $P(\varnothing )\to A$. Every such function defines a natural transformation by this formula, because restriction maps compose. This gives $\mathrm{Nat}(P,\Delta A)\cong \mathbf{Set}(P(\varnothing ),A)$.

For the second adjunction, a natural transformation $b:\Delta A\to P$ satisfies $b_U={\mathrm{res}}_{S,U}\circ b_S$, so it is uniquely determined by a function $A\to P(S)$. Every such function defines the remaining components by restriction, giving $\mathrm{Nat}(\Delta A,P)\cong \mathbf{Set}(A,P(S))$. Evaluation acts on presheaf morphisms componentwise, so these object assignments are indeed functors. The use of $P(\varnothing )$ is essential: a constant presheaf of sets has value $A$ there, whereas a sheaf has a singleton value there. No sheaf condition is imposed in this question.

Use Church numeralsExplicit lambda terms for the successor and zero test areDirect beta reduction gives $\mathsf{Succ}{c}_n{\equiv }_{\beta }{c}_{n+1}$, and the Church numeral zero test returns $\mathsf{T}$ exactly at zero. For the predecessor, use the Church pair operationsand setAfter $j$ iterations this pair holds $(c_j,c_{\max (j-1,0)})$: the initial pair gives the case $j=0$, and one step moves the old first coordinate into the second and increments the first. This proves the predecessor equation, including $\mathsf{Pred}{c}_0{\equiv }_{\beta }{c}_0$.

A total function is a lambda-definable function when a closed lambda term sends each tuple of numerals to the numeral of its value. Suppose $G,H$ represent total $g,h$, andFor lambda definition of primitive recursion by pair iteration, putOn numeral inputs, induction gives the pair $(c_j,c_{r(\mathbf{x},j)})$ after $j$ iterations. The base case is the equation for $G$, and the induction step is exactly the equation for $H$. Hence $R$ represents $r$, proving closure of total lambda-definable functions under primitive recursion.

For unbounded minimization, the careful closure statement is partial: if total $f(\mathbf{x},n)$ is lambda-definable, thenis a lambda-definable partial function, defined when a zero exists and otherwise undefined. It need not be total. With $F$ representing $f$, use a fixed-point combinator to defineEach test terminates because $f$ is total. Normal-order beta reduction chooses the return branch at the first zero and does not evaluate the discarded branch. If there is no zero, the required head computation passes through tests for all $n$, never exposing a numeral; standardization theorem for beta reduction rules out a hidden numeral reduct. For partial $f$, the usual mu operator is defined only if all earlier tests terminate and the successful test is zero. This stronger partial closure requires representatives that diverge on undefined inputs, or an explicit strict evaluator; the weaker condition of producing no numeral on undefined inputs alone must not be used as a test for convergence.

The zero function, successor, and projections have immediate numeral representations; function composition in recursion theory is implemented by substitution on total inputs. Thus every primitive recursive function is lambda-definable. The Kleene normal form theorem writes every partial recursive function as a total primitive-recursive decoding after a single minimization of a total primitive-recursive computation-history predicate. The search above consequently represents every partial recursive function, diverging when no successful history exists. Conversely a machine can effectively enumerate the beta reductions of a fixed representing lambda term on numeral inputs until it finds a Church numeral; distinct numerals cannot be beta-equivalent by the Church-Rosser theorem. It computes exactly the represented partial function. Therefore lambda-definable partial functions are precisely partial computable functions, and the total ones are precisely total computable functions.

This also gives the full partial minimization closure without assuming that an arbitrary weak representative diverges on every undefined input. Run its effective numeral evaluator successively at $n=0,1,2,\dots$; stop at the first zero, and run forever if an earlier evaluation fails to terminate or no zero is found. This is exactly the sequential-domain convention for the mu operator. The resulting partial computable function has a lambda representation by the construction just proved. Thus partial lambda-definable functions are closed under partial minimization; a minimized function belongs to the total family only when the required successful search exists for every input.