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For equally spaced nodes $x_0,x_0+h,\dots ,x_0+dh$, write $t=(x-x_0)/h$. The interpolating polynomial is $P(x)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{t}{j}\right){\Delta }_h^{j}f(x_0)$. A table of forward differences supplies its coefficients; for a polynomial of degree at most $d$, the formula reproduces the function exactly.

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.

In the monomial basis $\mathcal{B}=(1,x,x^2,x^3)$ of the complex vector space $V=\mathbb{C}[x{]}_{\le 3}$, the binomial theorem givesThe characteristic polynomial of this upper triangular matrix is the product of $t$ minus its diagonal entries, so ${\chi }_T(t)=(t-1{)}^4$.

For the minimal polynomial, set $\Delta =T-I$, the forward difference with step one. For a nonconstant polynomial of polynomial degree $d$, the binomial theorem shows that $f(x+1)-f(x)$ has polynomial degree $d-1$. Constant polynomials have zero forward difference. Consequently ${\Delta }^4=0$ on $V$. To prove that the nilpotency index is exactly four, computeIt follows thatIf $T$ were diagonalizable, its only eigenvalue would be $1$, and its diagonal matrix in an eigenbasis would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus $T$ is not diagonalizable. Equivalently, its minimal polynomial has a repeated root.

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.