Simple question about the Dirac delta function.

Question

I'm a bit confused by the dirac delta function.
It may sound quite silly, but i'm not quite sure about this.
Let's consider $p(x) = \frac{\delta(x-x_1)}{\delta(x-x_1) + \delta(x-x_2)}$ where $x_1$ and $x_2$ is real number and $x_1 \neq x_2$.

My simple question is then,
can we say $p(x)=1$ for $x=x_1$ and 0 for $x \neq x_1$??


What happen in the case $x_1 = x_2$?
can we say $p(x) = \frac{1}{2}$ for ${}^{\forall} x \in \Re$?

Thank you!

Answer 1

I would recommend to use a limit representation of the delta function for such purposes. For instance, one can use that: \begin{eqnarray} \delta(x-x_{0})=\lim_{\eta\to 0^{+}}\frac{1}{\pi}\frac{\eta}{(x-x_{0})^{2}+\eta^{2}} \end{eqnarray} although there are many more limiting cases of functions that reduce to the delta function (see for example the Gaussian case or the $\text{sinc}(x)$ function). Using this representation, you could work out the different limits $\to x_{1,2}^{\pm}$, where $\pm$ means approaching from left or right. The parameter $\eta$ eventually tends to $0$ and you can evaluate the limits.

Answer 2

The only way I can understand your definition of $p(x)$ is as the distribution solution of the equation $$ (\delta_{x_1} - \delta_{x_2})\,p(x) = \delta_{x_1} $$ In order to give a meaning to this equation, $p$ has to be continuous in $x_1$ and in $x_2$. Since $\delta_c(x)\, p(x) = \delta_c(x)\, p(c)$ $$ \delta_{x_1}\,p(x_1) - \delta_{x_2}\,p(x_2) = \delta_{x_1} $$ So, any function $p$ such that $p(x_1)=1$ and $p(x_2)=0$ works.

As a summary, $$ p(x) = \frac{\delta_{x_1}}{\delta_{x_1} - \delta_{x_2}} $$ does not defines a unique function, but any solution of this equation verifies $p(x_1)=1$.