Past exam of the approximation of functions course of the University of Oxford / 2011 / 1 / b / Solution

Assume $h>0$. The functional is the third forward difference: each forward difference lowers the polynomial degree by one, so it annihilates every polynomial of degree at most two. Directly, $X[1]=-1+3-3+1=0$, $X[x]=3h-6h+3h=0$, and $X[x^2]=3h^2-12h^2+9h^2=0$.

The Peano kernel theorem now gives $X[u]={\int }_0^{3h}K(t)u^{(3)}(t)dt$, where the Peano kernel of the third forward difference isHere $s_{+}=\max (s,0)$ is the positive-part truncated power function. The normalized quadratic cardinal B-spline isIts three polynomial pieces agree with the displayed kernel after rescaling, giving $\boxed{K(t)=h^2{B}_2(t/h)}$. Its knots are $0,h,2h,3h$ in the physical variable, and $K$ is positive throughout $(0,3h)$.

The kernel mass can be evaluated without integrating each piece: apply its representation to $u(x)=x^3/6$, whose third derivative is one. This gives ${\int }_0^{3h}K(t)dt=X[x^3/6]=h^3$. The weighted mean value theorem for integrals with a nonnegative weight therefore yieldsContinuity of $u^{(3)}$ and strict positivity of the weight in the interior place an attained weighted-average value in that open interval, including the constant-derivative case.