I was asked to "justify" the following formula with delta functions:
$\delta_{0}((x-a)(x-b))=\displaystyle\frac{(\delta_{0}(x-a)+\delta_{0}(x-b))}{|a-b|}$
Where we say $\delta_{0}(x-a)$ is the Dirac delta with concentration at $a$. However, I'm not sure on how to go about the proof for this statement. Without loss, one can assume that $b > a$, however I find that the result will be $\delta_{0}(x-a)+\delta_{0}(x-b)$, without the term $\frac{1}{|a-b|}$. Would in this case $\delta_{0}((x-a)(x-b))$ be the Dirac delta with concentration at $a,b$? How would one be able to justify the above formula (preferably using distributions and actions on test functions)?
Any help would be appreciated.
Where $f(x)$ is an arbitrary function, consider \begin{align} \int_{-\infty}^\infty \delta[g(x)]f(x)dx \end{align}
Assume that $g(x)$ has only one zero at $x=x_{0i}$ in the interval $(x_i,x_{i+1})$. Let's also assume $g'(x)>0$ on the interval (the case if g'(x)<0 should be handled separately). Then, the above integral is equivalent to
\begin{align} \int_{x_i}^{x_{i+1}} \delta[g(x)]f(x)dx \end{align}
Now let $u=g(x)$ such that $du = g'(x) dx$. Noting that $g(x_i) <g(x_{i0})=0 < g(x_{i+1})$, we find that
\begin{align} \int_{g(x_i)}^{g(x_{i+1})} \delta(u)\frac{f(g^{-1}(u))}{g'(g^{-1}(u))}du = \frac{f(g^{-1}(0))}{g'(g^{-1}(0))} = \frac{f(x_{0i})}{g'(x_{0i})} \end{align}
Strictly, I should be equating the delta function with a distribution in the above argument, arguing that I can extend the integration limits to infinity, and limiting the distribution until it achieves the form of a delta function. If we also considered the case where $g'(x)<0$ instead, we would find that in general, \begin{align} \int_{x_i}^{x_{i+1}} \delta[g(x)]f(x)dx = \frac{f(x_{0i})}{|g'(x_{0i})|} \end{align}
But what if $g(x)$ has multiple zeros, as in your case? Split up the integration limits into $N$ intervals labelled by $(x_i,x_{i+1})$ in a way that $g'(x)\ne0$ within each interval but may be zero on the boundaries. Since $g'(x)$ is either positive or negative in each interval, $g(x)$ has at most one zero within each interval. By the above argument, we can show
\begin{align} \sum_{i=1}^N \int_{x_i}^{x_{i+1}} \delta[g(x)]f(x)dx & =\sum_{i=1}^N \frac{f(x_{0i})}{|g'(x_{0i})|} \end{align}
One could say then, for a $g(x)$ with $N$ zeros, we can identify $$\delta[g(x)] = \sum_{i=1}^N \frac{\delta(x-x_{0i})}{|g'(x_{0i})|}$$
This matches what you're seeking. I hope this outline helps.