The B-Spline basis functions as defined by the Cox-DeBoor formula cannot, in general, be constructed
with convolution.
The convolution construction, as I'll explain below, only works for the special case when the knot vector is $\mathbb{Z}$
(i.e., the uniform knot vector with interval 1, sometimes called the cardinal spline).
To see this construction won't work in general, we can look at a simple example of a knot vector $0,1,3,4$.
The basis of the Cox-DeBoor recursive formula (the degree-0 basis functions) gives us a step function over
$[0,1]$, $[1,3]$ and $[3,4]$ (see for example here).
$$
N_{0,0}(u) = \left\{
\begin{array}{ll}
1 & 0 \leq u < 1 \\
0 & \text{otherwise} \\
\end{array}
\right\}
$$
$$
N_{1,0}(u) =
\left\{
\begin{array}{ll}
1 & 1 \leq u < 3 \\
0 & \text{otherwise} \\
\end{array}
\right\}
$$
$$
N_{2,0}(u) =
\left\{
\begin{array}{ll}
1 & 3 \leq u < 4 \\
0 & \text{otherwise} \\
\end{array}
\right\}
$$
The degree-1 basis functions, by the Cox-DeBoor formula, are then (the "triangular functions"):
$$
N_{0,1}(u) =
\left\{
\begin{array}{ll}
u & 0 \leq u < 1 \\
(3-u)/2 & 1 \leq u < 3 \\
0 & \text{otherwise} \\
\end{array}
\right\}
$$
$$
N_{1,1}(u) =
\left\{
\begin{array}{ll}
(u-1)/2 & 1 \leq u < 3 \\
4-u & 3 \leq u < 4 \\
0 & \text{otherwise} \\
\end{array}
\right\}
$$
However, convolving $N_{0,0}$ with $N_{1,0}$ will not give you $N_{0,1}$, nor will convolving $N_{1,0}$ with $N_{2,0}$.
To get a triangular function from $N_{1,0}$, you will need to convolve it with a step function of support 2, and that will not result in $N_{0,1}$ either (since the result is of support 4 and $N_{0,1}$ is of support 3).
The convolution construction of the cardinal B-Spline basis functions results from the B-Spline
derivative recursive formula (which itself results from Cox-DeBoor, see answers here and here for example).
The following development follows DeBoor's exposition in Section 10 of his paper.
Given the B-Spline recursive derivative formula:
$$
N_{i,p}'(u) = \frac{p}{u_{i+p}-u_i} N_{i,p-1}(u) - \frac{p}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u)
$$
We can integrate both sides and get (from the basic theorem of calculus):
$$
N_{i,p}(u) = \frac{p}{u_{i+p}-u_i} \int_{-\infty}^{u} N_{i,p-1}(t) dt -
\frac{p}{u_{i+p+1}-u_{i+1}} \int_{-\infty}^{u} N_{i+1,p-1}(t) dt
$$
When the knots are equally spaced (and only then) the coefficients cancel themselves, since
$u_{i+p}-u_i = u_{i+p+1}-u_{i+1}$.
Furthermore, if the equally spaced interval is unit length the coefficient is 1 (since $p=u_{i+p}-u_i$), and we get:
$$
N_{i,p}(u) = \int_{-\infty}^{u} N_{i,p-1}(t) - N_{i+1,p-1}(t) dt
$$
Now, when the knots are equally spaced $N_{i+1,p-1}$ is just a translation of $N_{i,p-1}$,
(indeed all the basis functions are just translations of themselves) so:
$$
N_{i+1,p-1}(t+1) = N_{i,p-1}(t)
$$
Therefore, we can write:
$$
N_{i,p}(u) = \int_{-\infty}^{u} N_{i,p-1}(t) dt - \int_{-\infty}^{u-1} N_{i,p-1}(t) dt
= \int_{u-1}^{u} N_{i,p-1}(t) dt
$$
The right-hand side can be re-written as:
$$
N_{i,p}(u) = \int_{-\infty}^{\infty} N_{0,0}(u-t) N_{i,p-1}(t) dt
$$
which is
the convolution formula of $N_{i,p-1}$
with $N_{0,0}$.
Thus,
$$
N_{i,p} = N_{0,0}*N_{i,p-1}
$$
To conclude, the convolution formula is related to the Cox-DeBoor formula through the derivative
equation, but only for uniform, unit-support knot vectors.
For non-uniform knot vectors the basis functions will not be a translation of each other, nor will the coefficients of the integrals be the same.
Furthermore, even for uniform
knot vectors with support $h \neq 1$ the convolution formula won't work as-is
and will require a correction scalar multiplication by $1/h$ (since the coefficient terms will be $1/h$).
