= 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> is
$$
K(t)=\frac12\left[3(h-t)_+^2-3(2h-t)_+^2+(3h-t)_+^2\right]
=\boxed{\begin{cases}
t^2/2,&0\leq t\leq h,\\
(-2t^2+6ht-3h^2)/2,&h\leq t\leq2h,\\
(3h-t)^2/2,&2h\leq t\leq3h.
\end{cases}}
$$
Here $s_+=\max(s,0)$ is the positive-part <truncated power function>. The normalized <quadratic cardinal B-spline> is
$$
B_2(s)=\frac12\left[s_+^2-3(s-1)_+^2+3(s-2)_+^2-(s-3)_+^2\right].
$$
Its three <polynomial> pieces agree with the displayed kernel after rescaling, giving $\boxed{K(t)=h^2B_2(t/h)}$. 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 yields
$$
\boxed{X[u]=h^3u^{(3)}(\xi),\qquad0<\xi<3h.}
$$
Continuity 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.
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