Peano kernel ($K(\theta )$)

The Peano kernel of a linear error functional $L$ that annihilates all polynomials of degree at most $k$ isIt represents the error by integrating $K$ against the $(k+1)$st derivative.

The Peano kernel theorem writes a derivative-based error functional as $L(u)=\int K(t)u^{(r)}(t)dt$, so its bound in the displayed derivative seminorm is at most the L1 norm of the kernel. This is the sharp constant: continuous derivatives can approximate the sign of $K$ in the weighted integral, and integrating them $r$ times produces admissible test functions. If $K$ has one sign, a constant $r$th derivative already attains equality; in particular $\int K=L[x^r/r!]$. This provides a direct exact check on proposed error constants before estimating individual polynomial pieces.

For $h>0$, the third forward difference $L_{h}u=-u(0)+3u(h)-3u(2h)+u(3h)$ annihilates polynomials of degree at most two. Its Peano kernel is $K_h(t)=\frac{1}{2}[3(h-t{)}_{+}^2-3(2h-t{)}_{+}^2+(3h-t{)}_{+}^2]$ for $0\le t\le 3h$, extended by zero outside that interval. It is $h^2$ times the quadratic cardinal B-spline evaluated at $t/h$, is positive on $(0,3h)$, and has integral $h^3$. The Peano kernel theorem therefore represents $L_{h}u$ as a positive weighted average of the third derivative: $L_{h}u=h^3{u}^{(3)}(\xi )$ for some interior $\xi$ when that derivative is continuous. The kernel is the smooth error weight hidden inside a discrete difference.

For $L(f)=f^{\prime \prime }(0)-f(-1)+2f(0)-f(1)$ on $C^3[-1,1]$, quadratic polynomials are annihilated. The Peano kernel theorem gives $L(f)={\int }_{-1}^{1}K(t)f^{\prime \prime \prime }(t)dt$, with the displayed kernel; its value at zero does not affect the integral. Its absolute integral is $1/3$, proving $|L(f)|\le \Vert f^{\prime \prime \prime }{\Vert }_{\infty }/3$. The kernel changes sign, consistent with exactness also on cubic polynomials despite the estimate using only three derivatives.

For $L(f)=-5f(0)/3+4f(1/2)/3+f(1)/3-f^{\prime }(1/3)$ on $C^3[0,1]$, quadratic polynomials are annihilated. Taylor's formula with integral remainder gives $L(f)={\int }_0^{1}K(t)f^{\prime \prime \prime }(t)dt$ with $K(t)=2(1/2-t{)}_{+}^2/3+(1-t{)}^2/6-(1/3-t{)}_{+}$. This Peano kernel is nonnegative throughout the interval and integrates to $1/36$, giving the sharp bound; $f(x)=x^3/6$ attains equality. The derivative is at an interior point, so this is distinct from an endpoint differentiation formula.

If $\lambda (f)=k{!}^{-1}\int K{f}^{(k+1)}$, the best unrestricted derivative-norm bounds are $|\lambda (f)|\le \Vert K{\Vert }_{\infty }\Vert f^{(k+1)}{\Vert }_1/k!$ and $|\lambda (f)|\le \Vert K{\Vert }_1\Vert f^{(k+1)}{\Vert }_{\infty }/k!$. Continuous concentrated bumps approach the first constant, and continuous approximations to $\mathrm{sgn}K$ approach the second. For the three-point asymmetric second-derivative formula, these constants are $5/6$ and $7/9$.

For $L(f)=f^{\prime }(0)+\frac{3}{2}f(0)-2f(1)+\frac{1}{2}f(2)$ on $C^3[0,2]$, the Peano kernel isThe Taylor expansion with integral remainder gives $L(f)={\int }_0^{2}K(t)f^{(3)}(t)dt$, because $L$ annihilates polynomials of degree at most two. Since $K\ge 0$ and ${\int }_0^{2}K=1/3$, the sharp bound is $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. Equality holds for $f(x)=x^3/6$.