Direct evaluation gives $X[1]=X[x]=X[x^2]=0$ and $X[x^3]=-13/16$. Hence the linear functional vanishes on every polynomial of degree at most two and is nonzero on every genuine cubic polynomial.

The Peano kernel theorem with order three gives $X[u]={\int }_{-1}^{1}K(t)u^{\prime \prime \prime }(t)dt$, whereThe first term comes from differentiating $(x-t{)}_{+}^2/2$ at $x=1/4$. On the integration interval this simplifies toIn particular $K\le 0$. Its L1 norm is $-\int K=-X[x^3/6]=13/96$, so the sharp bound isThe printed constant $1/12$ is false for the printed functional. The cubic $u=x^3/6$ has $\Vert u^{\prime \prime \prime }{\Vert }_{\infty }=1$ but $|X[u]|=13/96>1/12$, and it attains the corrected bound. This is an exact Peano kernel norm calculation, not an error from numerical quadrature.

Use the standard meaning of a cubic spline: a $C^2$ function that is cubic on each unit interval. If it vanishes on both exterior intervals, continuity of the function and its first two derivatives gives zero endpoint jets at zero and at $k$.

Such a spline can be represented through its third-derivative jumps asFor $x\ge k$, this must be the zero polynomial. Expanding in powers of $x$ yields the four moment conditionsIf $k\le 3$, the columns of this Vandermonde system are linearly independent, so every $c_j=0$. No nonzero spline has such short support.

For $k=4$, a nonzero null vector is $(1,-4,6,-4,1)$, since the fourth finite difference annihilates every cubic. Thus the Cardinal cubic B-spline isIts knot values at $0,1,2,3,4$ are $0,1/6,2/3,1/6,0$, whose sum is one. It is nonzero and supported on $[0,4]$, proving the minimal value $k=4$. Equivalently, on $[0,1]$ it is $x^3/6$, on $[1,2]$ it is $[-3x^3+12x^2-12x+4]/6$, and the other half follows by symmetry $\varphi (4-x)=\varphi (x)$.

By definition, a best uniform approximation $p^{\ast }\in P_n$ minimizes $\Vert u-p{\Vert }_{\infty }$ over $p\in P_n$. Write $E=\Vert u-p^{\ast }{\Vert }_{\infty }$. The equioscillation theorem says that, if $E>0$, optimality is equivalent to the existence of $n+2$ ordered points with signed errorsThe absolute-value bars on the left of the printed formula must be omitted for this alternating-sign statement. A nonnegative absolute value cannot equal a negative $-E$.

For sufficiency, suppose a polynomial $p$ had strictly smaller error. At the alternating points, $p-p^{\ast }$ must have the same sign as $u-p^{\ast }$, since otherwise the new error would have magnitude at least $E$. The difference polynomial therefore changes sign in all $n+1$ successive gaps and has at least $n+1$ distinct roots, impossible for a nonzero element of $P_n$.

For necessity, suppose that no $n+2$ alternating extremal points exist. On the compact extremal set $K=\{x:|u(x)-p^{\ast }(x)|=E\}$, group the signs of the error into consecutive blocks. The positive and negative extremal sets are disjoint compact sets, so sign changes are separated by positive distances; the number of blocks is finite. The absence of an $n+2$ alternating selection means there are at most $n+1$ blocks. Place a root in each gap between successive blocks and choose the overall sign of their product $q$ to agree with the error on $K$. Then $\mathrm{deg}q\le n$ and $q$ is nonzero throughout $K$.

By compactness and continuity, on a neighborhood of $K$ the signs still agree and $|q|$ is bounded below. A sufficiently small positive $\eta$ reduces the error there when $p^{\ast }$ is replaced by $p^{\ast }+\eta q$. On the compact complement of this neighborhood the old error is below $E$ by a fixed positive margin, and choosing $\eta$ smaller preserves that margin. This gives a strictly better approximation, a contradiction. If $E=0$, the target already lies in $P_n$ and optimality is immediate.

For $u(x)=x^{1/3}$, the chord is $x$. Concavity gives $x^{1/3}-x\ge 0$, with maximum at $x_{\ast }=1/(3\sqrt{3})$ and maximum value $2/(3\sqrt{3})$. Shift the chord up by half that amount. The errors at $0,x_{\ast },1$ are respectively $-E,+E,-E$, so

Put $x=1-t$. For $0<k\le 1$,whereas the constrained rational interpolant expands asMatching the constant term gives $a=1+b$, and matching the coefficient of $t$ gives $1/(1+b)=k$. Thus $a=1/k$, $b=(1-k)/k$, andThis is an endpoint-constrained rational approximation to a power: it matches value and derivative at $x=1$ and also enforces $r(0)=0$ for $k>0$.

There is a terminology qualification. A standard degree-$(1,1)$ Padé approximant about $t=0$ matches through $t^2$, but the displayed constrained expression has quadratic coefficient $-k(1-k)$ rather than $-k(1-k)/2$. The unconstrained Padé approximant is insteadThe source's requested formula is obtained by the two stated endpoint matching conditions within its restricted family, rather than by full Padé matching. At $k=1$ both reduce to $x$. At $k=0$, the requested expression is one for $x>0$ with a removable value at zero; finite $a,b$ in the original restricted form do not represent that constant function. The endpoint is understood by this limit.

The best approximation in the $L^2$ norm is the orthogonal projection onto $\mathrm{span}\{{\varphi }_0,{\varphi }_1\}$. ThereforeUsing ${\int }_0^1{x}^{k}dx=1/(k+1)$,It follows thatThe error is orthogonal to every linear polynomial, so the Pythagorean identity proves optimality and uniqueness. At $k=0$, this gives $p_1=1$; at $k=1$, it gives $p_1=x$. Both are exact because the target itself then belongs to $P_1$.

The monic degree-zero polynomial is ${\varphi }_0=1$. Orthogonality of $x-a$ to one gives $a={\int }_0^{1}xdx=1/2$, so ${\varphi }_1=x-1/2$.

The relevant integrals areInsert these into the three-term recurrence for monic orthogonal polynomials:ThusTheir mutual orthogonality can also be checked directly by integrating on $[0,1]$.

The monic orthogonal polynomials ${\varphi }_0,\dots ,{\varphi }_{r+1}$ form a basis of $P_{r+1}$, ordered by degree. Since $x{\varphi }_r$ is monic of degree $r+1$,For $j\le r-2$, symmetry of multiplication by $x$ gives $(x{\varphi }_r,{\varphi }_j)=({\varphi }_r,x{\varphi }_j)=0$, because $\mathrm{deg}(x{\varphi }_j)<r$. Only $a_r$ and $a_{r-1}$ remain. Taking inner products with ${\varphi }_r$ yields $a_r=({\varphi }_r,x{\varphi }_r)/({\varphi }_r,{\varphi }_r)$.

Similarly,since $x{\varphi }_{r-1}-{\varphi }_r$ has degree at most $r-1$. Rearranging proves the three-term recurrence for monic orthogonal polynomials:

Let $t_1,\dots ,t_m$ be the distinct roots in $(0,1)$ at which ${\varphi }_r$ changes sign, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Both ${\varphi }_r$ and $q$ change sign at each such point, while roots of even multiplicity do not change sign. Consequently ${\varphi }_{r}q$ has one constant sign on $(0,1)$ except at its zeros, and is not identically zero. Its integral cannot vanish.

If $m<r$, however, orthogonality of the monic orthogonal polynomial ${\varphi }_r$ to every polynomial of lower degree gives $({\varphi }_r,q)=0$, a contradiction. Hence $m\ge r$. Since $\mathrm{deg}{\varphi }_r=r$, there must be exactly $r$ such roots, each of multiplicity one. They exhaust the degree, leaving no complex, exterior, or endpoint roots. Thus all roots are simple and lie strictly in $(0,1)$, which proves the requested weaker inclusion in $[0,1]$. This is the sign-change proof of the zeros of orthogonal polynomials theorem.

Use the Chebyshev polynomial identities $T_0=1$, $T_1=x$, and $T_{j+1}=2x{T}_j-T_{j-1}$. Induction shows that $T_j$ has degree $j$ with leading coefficient $2^{j-1}$ for $j\ge 1$. The trigonometric definition also givesThereforeThe leading coefficient is $2^{1-n}{n}^{-1}\cdot n{2}^{n-1}=1$, and the degree is $n+1$. The endpoints are roots through $x^2-1$; the other $n-1$ points $\cos (s\pi /n)$ are roots through $T_n^{\prime }$. They are distinct, so this monic polynomial is exactly ${\pi }_n$.

Differentiating the product ${\pi }_n(x)=(x-x_s){\pi }_{n,s}(x)$ at $x_s$ gives ${\pi }_n^{\prime }(x_s)={\pi }_{n,s}(x_s)=w_s^{-1}$. For an interior node, differentiate the trigonometric form using $d/dx=-(\sin \theta {)}^{-1}d/d\theta$:At ${\theta }_s=s\pi /n$, this is $2^{1-n}n(-1{)}^s$. At either endpoint, taking the limit makes the two terms equal, doubling the value: ${\pi }_n^{\prime }(x_s)=2^{2-n}n(-1{)}^s$ for $s=0,n$.

After removing the common factor $2^{n-1}/n$, the barycentric interpolation formula therefore usesThese Chebyshev-Lobatto node weights differ in magnitude by only a factor of two.

For uniformly spaced nodes,There are $n-s$ negative factors, coming from $j>s$. Thus the weights fixed by part (b) satisfy this formula exactly. The printed sign $(-1{)}^s$ agrees for even $n$; for odd $n$ it differs by the common factor $-1$. Rescaling all weights by $(-1{)}^n$ gives the printed convention without changing the barycentric interpolation formula.

For even $n$, the required magnitude isStirling's formula givesThus central weights are exponentially larger in magnitude than endpoint weights. Their alternating signs can cause severe cancellation in polynomial interpolation at equispaced nodes; they reflect the unfavorable high-degree interpolation geometry associated with Runge's phenomenon. A barycentric evaluation formula improves the implementation but does not remove the intrinsic sensitivity or guarantee convergence for every smooth function. The comparison with the bounded relative weights at Chebyshev-Lobatto nodes is decisive.

With $w_s=1/{\pi }_{n,s}(x_s)$, the Lagrange interpolation polynomial can be written, away from its nodes,Apply the same formula to the constant data $u_s=1$. Uniqueness of polynomial interpolation gives ${\sum }_s{\ell }_s=1$, and henceDivide the two identities to obtain the barycentric interpolation formulaAt $x=x_r$, use the continuous limiting value $u_r$. Multiplying every weight by the same nonzero constant cancels between numerator and denominator, so only relative weights are needed in computation.

Define the Lagrange cardinal polynomialsDistinct nodes make every denominator nonzero. At $x_s$ these polynomials satisfy ${\ell }_r(x_s)={\delta }_{rs}$, so the Lagrange interpolation polynomial isIt has degree at most $n$ and takes the prescribed data values. If two such polynomials existed, their difference would have $n+1$ distinct roots but degree at most $n$, forcing that difference to be zero. This also proves uniqueness.

Write $u_h^m={\sum }_i{U}_i^m{\phi }_i$ and define the mass matrix and reaction-diffusion stiffness matrixThen each step isBoth component matrices are symmetric, so $A$ is symmetric. For a nonzero coefficient vector $v$, its finite element function $v_h$ is nonzero andThus $A$ is also positive definite and every update has a unique solution. Because nonneighboring hats have disjoint interior supports, the matrix is tridiagonal in nodal order.

Test the discrete equation with $v_h=u_h^{m+1/2}$. The identity $(a-b,a+b)=\Vert a{\Vert }^2-\Vert b{\Vert }^2$ yieldsSince $e^x\ge 1$ on the interval, $a(v,v)\ge \Vert v{\Vert }_{H^1}^2$. Therefore the requested inequality holds:Discarding the nonnegative second term and iterating proves $\Vert u_h^m{\Vert }_2\le \Vert u_h^0{\Vert }_2$ for every positive $\Delta t$, with no restriction relating it to $h$. This is unconditional $L^2$ stability; the initial projection also satisfies $\Vert u_h^0{\Vert }_2\le \Vert u_0{\Vert }_2$.

Use the full continuous piecewise linear space $V_h\subset H^1(0,1)$, including endpoint basis functions, because the boundary conditions are Neumann. DefineChoose $u_h^0$ as the $L^2$ projection of $u_0$ onto $V_h$. For each time step seek $u_h^{m+1}\in V_h$ withThis is the conforming Crank-Nicolson method. The reaction coefficient and the diffusion are both treated at the midpoint; integration by parts has removed the boundary flux through the homogeneous Neumann condition.

Compute $C$ from part (b), solve the finite element equations, and calculate the element indicators ${\eta }_i^2=h_i^4\Vert R_h{\Vert }_{L^2(I_i)}^2$. Let $\eta =({\sum }_i{\eta }_i^2{)}^{1/2}$. Stop when $C\eta <\mathrm{TOL}$; the proved reliability bound then certifies the requested true-error tolerance.

Otherwise mark elements carrying a chosen fraction of the total indicator, bisect them, and repeat. To ensure termination as well as a valid stopping test, include a uniform-refinement safeguard, for example bisect every element on every second unsuccessful pass. This preserves adaptive refinement on the other passes while forcing $h_{\max }\to 0$.

Here termination can be justified directly. On each element $R_h=f+p^{\prime }{u}_h^{\prime }-q{u}_h$, so its global $L^2$ norm stays bounded if $\Vert u_h{\Vert }_{H^1}$ stays bounded. That stability follows from coercivity after subtracting the fixed lift one at the right endpoint. HenceThe safeguarded algorithm therefore eventually satisfies the strict tolerance. Reliable numerical evaluation of the residual integrals, or an upper bound for their quadrature errors, is part of using the certificate in an implementation.

Let $p_{\ast }=\min p>0$, $q_{\ast }=\min q>0$, $q^{\ast }=\max q$, and $P_1=\Vert p^{\prime }{\Vert }_{\infty }$. Testing the dual equation with $z$ givesConsequently $\Vert z{\Vert }_2\le \Vert e{\Vert }_2/q_{\ast }$ and $\Vert z^{\prime }{\Vert }_2\le \Vert e{\Vert }_2/\sqrt{p_{\ast }{q}_{\ast }}$. From $p{z}^{\prime \prime }=qz-e-p^{\prime }{z}^{\prime }$,On an element of length $h_i$, the interpolation error vanishes at both endpoints and its derivative has zero mean. Two Poincaré inequalities giveApply Cauchy-Schwarz to the residual identity, with factors $h_i^2$ and $h_i^{-2}$:If the error is nonzero, cancel its norm after the regularity estimate; if it is zero the conclusion is immediate. One explicit reliable constant isThis proves the requested dual residual bound entirely in terms of the coefficient functions.

Put $e=u-u_h$ and $V=\{v\in H^1(0,1):v(1)=0\}$. The primal solution and its finite element approximation lie in the same affine space with value one at the right endpoint, so $e\in V$. UseDefine the dual solution byEquivalently $z\in V$ satisfies $a(v,z)=(e,v)$ for every $v\in V$. Positivity of $p,q$ gives a unique solution, and one-dimensional regularity gives $z\in H^2$.

Then $\Vert e{\Vert }_{L^2}^2=a(e,z)$. The nodal interpolant $I_{h}z$ belongs to the homogeneous finite element test space, so Galerkin orthogonality gives $a(e,I_{h}z)=0$. Set $w=z-I_{h}z$. Using the primal weak equation,Integrate the derivative term separately on each element. Since $w$ vanishes at every mesh node, all endpoint flux and flux-jump terms vanish. ThereforeThe absence of jump terms here follows specifically from one-dimensional nodal interpolation; it is not an assumption that the finite element flux itself is continuous.

Take $V_h$ to be all continuous functions affine on each triangle of the stated negative-diagonal mesh. No boundary value is fixed, because the Neumann condition is natural. The discrete problem isIts matrix is symmetric positive definite: the reaction contribution $\int u_h{v}_h$ removes the constant null mode that would occur for pure Neumann Poisson equations. Thus the solution exists uniquely.

For every $v_h\in V_h$, expand the quadratic functional about $u_h$ and use the discrete weak equation to remove its linear term:The continuous solution minimizes $J$ over the larger space $H^1$, so

For smooth $u$, Green's first identity rewrites the weak equation asCompactly supported test functions yield $-\Delta u+u=f$. Since the test space is the full $H^1(\Omega )$ and has unrestricted boundary trace, the remaining identity gives the homogeneous Neumann condition.

For a merely weak solution on a polygon, the normal derivative is understood as its weak flux trace. Here $\nabla u\in L^2$ and $\mathrm{div}\nabla u=u-f\in L^2$, so this trace is defined through Green's formula. It vanishes by the same identity. Thus the natural boundary condition is ${\partial }_{n}u=0$, without assuming global $H^2$ regularity on an arbitrary polygon.