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For the Peano kernel theorem, the integral Taylor remainder isHere $u_{+}=\max (u,0)$. The error functional must be linear, vanish on every polynomial of degree at most $k$, and permit its action to be interchanged with this remainder integral. The last regularity condition holds for the finite combinations of point evaluations and derivative evaluations used below, by applying the corresponding differentiated remainder formula. Under these conditions the polynomial terms vanish, yielding

For the three-point asymmetric second-derivative formula, choose the error convention $\lambda (f)=f^{\prime \prime }(2)-L(f)$. Exactness on $1,x,x^2$ giveshenceApplying $\lambda$ to $(x-\theta {)}_{+}^2$, for almost every $\theta \in [0,3]$, gives the Peano kernelThe value at $\theta =2$ is immaterial to the integral. The sketch rises from zero through $2/3$ at one to the left limit $5/3$ at two, jumps down to the right limit $-1/3$, and then rises to zero at three.

The sharp error constants for a Peano kernel are the corresponding integral-operator norms. HereAs $k!=2$,Indeed, $|\lambda (f)|\le (\Vert K{\Vert }_{\infty }/2)\Vert f^{(3)}{\Vert }_1$ and $|\lambda (f)|\le (\Vert K{\Vert }_1/2)\Vert f^{(3)}{\Vert }_{\infty }$. They are best possible even for $f\in C^3$: for $r$, choose continuous nonnegative third derivatives concentrated just left of two with unit integral; for $s$, approximate the sign of $K$ by continuous functions of supremum norm one. Integrating any such third derivative three times supplies an admissible $f$, and the resulting ratios approach the stated constants.

For $f(x)=x^3$, $L(f)=-1+27/3=8$, while $f^{\prime \prime }(2)=12$, so $\lambda (f)=4$. Also $\Vert f^{(3)}{\Vert }_1=18$ and $\Vert f^{(3)}{\Vert }_{\infty }=6$. The two bounds explicitly read

The Peano kernel theorem says that if a continuous linear error functional $L$ annihilates all polynomials of degree less than $r$ and permits the required interchange with integration, then for $f\in C^r[a,b]$,It follows by applying $L$ to the integral-remainder Taylor formula with integral remainder. Here choose $r=2$ andThis functional annihilates constants and linear functions (indeed quadratics too), and its Peano kernel isThus ${\int }_0^3|K|d\theta =1/2$ and $\max |K|=1/3$. The integral representation givesFor the first inequality, continuous functions $f^{\prime \prime }$ with $\Vert f^{\prime \prime }{\Vert }_{\infty }\le 1$ can approximate $\mathrm{sgn}K$ away from its zeros, making $L(f)$ arbitrarily close to $\int |K|$. Such derivatives integrate twice to admissible $C^2$ functions, proving no smaller $\alpha$ works. For the second, use continuous $f^{\prime \prime }$ of $L^1$ norm one concentrated near $\theta =1$ or $2$, where $|K|=1/3$, with the sign of $K$ there. This makes $|L(f)|$ approach $1/3$, proving optimality of $\beta$.

For $f=x^3$, $L(f)=2$, $\max |f^{\prime \prime }|=18$ and $\Vert f^{\prime \prime }{\Vert }_1=27$. Both right-hand bounds equal nine, so both inequalities hold.

Let $x_1,\dots ,x_k$ be the distinct interior roots of $p_n$ where its sign changes, and put $q(x)={\prod }_{j=1}^k(x-x_j)$. If $k<n$, orthogonality gives ${\int }_a^{b}w{p}_{n}q=0$. But $p_{n}q$ has constant sign on $(a,b)$, apart from its isolated zeros, because every sign change has been canceled. It is not identically zero, and the positive weight makes its integral nonzero, a contradiction. Thus $k\ge n$. Degree $n$ forces exactly $n$ distinct simple roots, all inside $(a,b)$.

For distinct nodes, let ${\ell }_i(x)={\prod }_{j\ne i}(x-c_j)/(c_i-c_j)$. Lagrange interpolation polynomial gives $f={\sum }_{i}f(c_i){\ell }_i$ for every polynomial of degree at most $n-1$. Integrating this identity proves exactness with $b_i={\int }_a^{b}w{\ell }_i$.

For the Gaussian quadrature nodes, divide any $f\in {\mathbb{P}}_{2n-1}$ as $f=q{p}_n+r$, with $q,r\in {\mathbb{P}}_{n-1}$. The integral of $q{p}_n$ vanishes by orthogonality, and its nodal values vanish because $p_n(c_i)=0$. The remaining polynomial $r$ is integrated exactly by the preceding interpolation result. HenceIts weights are positive: exactness applied to ${\ell }_i^2$ gives $b_i={\int }_a^{b}w{\ell }_i^2>0$.

Define the error functional $L(f)={\int }_a^{b}w(x)f(x)dx-{\sum }_i{b}_{i}f(c_i)$. It annihilates polynomials of degree at most $2n-1$. For $f\in C^{2n}[a,b]$, the Peano kernel theorem giveswhere a formal Peano kernel with the normalization made explicit isNo evaluation of $K$ is needed. The supplied inner product presupposes an integrable positive weight on the finite interval, so the integral and point-evaluation error functional is bounded on continuous functions.

Repeated integration by parts inproduces the first $k+1$ terms of the Taylor polynomial and leaves $f(x)$, proving the integral remainder in Taylor theoremApply the linear functional ${\lambda }_k$ to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

For the midpoint quadrature rule on $[0,1]$, $e[1]=e[x]=0$, so $k=1$. Its Peano kernel isFor $f(x)=x^2$, the formula givesagreeing with ${\int }_0^1{x}^{2}dx-(1/2{)}^2=1/3-1/4$.

The Peano kernel theorem says that if a continuous linear functional $L$ annihilates polynomials of degree below $m$, thenHere $m=3$ andApplying $L$ to $(x-t{)}_{+}^2/2$ givesHenceThe smallest candidate is therefore

If the bound is to hold with finite $c$, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ givesSolving,This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional $L$ annihilates all polynomials of degree below $r$, then for $f\in C^r[a,b]$,

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,The integral equals $e(q)$ for $q(x)=(x+1{)}^4/24$, since $q^{(4)}=1$. Now $q^{\prime \prime }(-1)=0$ andThereforeEquality is attained by $q$, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

The integral form of the Taylor theorem about the left endpoint iswhere $q_{k-1}\in {\mathcal{P}}_{k-1}$. Apply the order-$k$ divided difference at $x_i,\dots ,x_{i+k}$. The polynomial term vanishes, and linearity permits interchange with the integral:By the definition $M_i(t)=k[x_i,\dots ,x_{i+k}](\cdot -t{)}_{+}^{k-1}$,This is the Peano kernel theorem for the divided-difference functional.

Matching the monomials $1,x,x^2$ gives $f^{\prime \prime }(2)\approx (2/3)f(0)-f(1)+(1/3)f(3)$. With error defined as the true derivative minus this approximation, its Peano kernel is $K(\theta )=21_{\{\theta <2\}}+(1-\theta {)}_{+}^2-(3-\theta {)}^2/3$ on $[0,3]$. The derivative evaluation causes a jump at two. Applying the Peano kernel theorem gives $\lambda (f)=\frac{1}{2}{\int }_0^{3}K{f}^{(3)}$.