I'm following a course on image processing and we have the following statement
$$ w = w_1 * w_2 = w_1 \cdot w_2 $$ $$w_1 \in R_{m \times 1}, w_2 \in R_{ 1 \times n }$$
How can one prove this with the formula for 2D discrete convolution ?
$$(w * f)(x,y) = \sum_{s=-a}^a \sum_{t=-b}^b w(s, t) \dot f(x-s, y-t) $$
First, you need to get the 2D discrete convolution correctly:
$$ (w*f)(x,y) = \sum_s\sum_t w(s,t)f(x-s,y-t)$$
where $s,t$ ranges are deduced from the dimensions of $w$ and $f$. In your case these dimensions are $m\times 1$ and $1\times n$ respectively. If we use 0-based indices than we get $t = 0$ and $x-s=0 \longrightarrow s = x$. Thus the convolution for this special case becomes
$$ (w*f)(x,y) = \sum_{s=x}\sum_{t=0} w(s,t)f(x-s,y-t) = w(x,0)f(0,y) $$ i.e., the whole sum has only one element. That element is the same as $(w\cdot f)_{x,y}$ where $w\cdot f$ is the matrix product of $w$ and $f$.