= Solution
In a convenient cardinal parameter normalization, take the unit $m$-cube and project by the sum of its coordinates. Its pushforward density is the <unit-direction univariate box spline>
$$
B_m(x)=\int_{[0,1]^m}\delta\!\left(x-\sum_{r=1}^m u_r\right)du_1\cdots du_m.
$$
Equivalently, with $B_1=\mathbf1_{[0,1)}$ at its discontinuity points,
$$
\boxed{B_m=B_1*\cdots*B_1,\qquad B_m(x)=\int_0^1B_{m-1}(x-u)\,du.}
$$
Here there are $m$ <convolution> factors; some conventions instead index the <spline> by its degree $m-1$. The half-open convention for $m=1$ assigns knot values consistently and does not change the pushforward measure.
\b[Support and positivity.] A sum of $m$ numbers in $[0,1]$ lies in $[0,m]$, so the closed support is finite. For $0<x<m$, the interval of $u\in(0,1)$ with $0<x-u<m-1$ has positive length. Induction in the <convolution> recurrence therefore gives $B_m(x)>0$ in the interior. For $m\geq2$, $B_m(0)=B_m(m)=0$: strict positivity includes the open support interval, not its endpoints.
\b[<Partition of unity>.] The translates of $B_1$ cover the line exactly once under the half-open convention. Nonnegativity permits interchange of sum and integral, and induction gives
$$
\sum_{j\in\mathbb Z}B_m(x-j)
=\int_0^1\sum_{j\in\mathbb Z}B_{m-1}(x-u-j)\,du
=1.
$$
Thus $\boxed{\sum_jB_m(x-j)=1}$, a locally finite sum. Each $B_m$ also has integral one, although <partition of unity> does not mean its maximum value is one.
For $m\geq2$, repeated integration gives the explicit truncated-power form
$$
B_m(x)=\frac1{(m-1)!}\sum_{j=0}^m(-1)^j\binom mj(x-j)_+^{m-1}.
$$
On each unit knot interval this is a <polynomial> of degree at most $m-1$, and on $(0,1)$ it is $x^{m-1}/(m-1)!$, so the degree is attained. Each truncated power has $m-2$ continuous <derivatives>; the <derivative> of order $m-1$ jumps at the first knot. Therefore the dependence on the projected box dimension is
$$
\boxed{\text{degree }m-1,\qquad \text{continuity }C^{m-2},\qquad
\text{support width }m\text{ in unit knot intervals}.}
$$
For $m=1$ the pieces are constants and the function is discontinuous at the knots. If geometric width is measured under orthogonal projection onto the unit diagonal instead, the projected coordinate is $x/\sqrt m$, so the physical support width is $\sqrt m$; its unit-mass density is $\sqrt m B_m(\sqrt m\,x)$. These are different scalings of the same cardinal <spline>, not different <continuity> or degree claims.
For a bivariate <box spline>, choose a spanning direction <matrix> $\Xi=(\xi_1,\ldots,\xi_m)$ in the plane and project the unit cube by $u\mapsto\Xi u$. Define its normalization by
$$
\int_{\mathbb R^2}M_\Xi(x)\varphi(x)\,dx
=\int_{[0,1]^m}\varphi(\Xi u)\,du.
$$
The support is the <zonotope> $\sum_r[0,1]\xi_r$, and the <polynomial> pieces generally have total degree $m-2$. The grid comes from the chosen direction families and their lattice translates. Two independent directions, with repetitions, give tensor-product <B-splines> on a parallelogram grid. Three families such as $e_1,e_2,e_1+e_2$ give a three-direction triangular grid; an affine change of basis can make it the usual equilateral triangular grid. The four families $e_1,e_2,e_1+e_2,e_1-e_2$ give the criss-cross square grid and the <quadratic four-direction box spline>. Changing the direction <matrix> and repeating directions therefore changes the grid, support and smoothness while retaining the box-projection construction.
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