Let $f$ be the distribution of a normal variable $\mathcal{N}(0,1)$, ie $$ f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$
I have to solve the equation:
$$x+y = f(x) - f(y)$$
I was working with mathematica and it gives me that the solution is $x=-y$. I would very happy if it's true, but I have no idea to prove that it is a necessary condition. Any advice? Thank you in advance!
Since $f$ is even, your question is equivalent to solving $x-y=f(x)-f(y)$, for $x,y\in\mathbb{R}$. Assume $x\neq y$. Then by mean value theorem, there exists some $t$ in the interval $]x,y[$, such that $$\frac{f(x)-f(y)}{x-y}=\frac{-te^{\frac{-t^2}{2}}}{\sqrt{2\pi}}.$$ We are hence reduced to prove that the function $$t\mapsto \phi(t)= te^{\frac{-t^2}{2}}+\sqrt{2\pi}$$ does not vanish on $\mathbb{R}$. One way to do that is to study its derivative given by $\phi '(t)=(1-t^2)e^{\frac{-t^2}{2}}$, and then notice that $\phi$ has a minima in $\mathbb{R}$ in $t=-1$ and $\phi(-1)=\sqrt{2\pi}-e^{\frac{-1}{2}}>0$ (no mathematica is needed here). This concludes the proof.
Alternative way: we solve $f(x)-f(y)-x+y=0$.
Fix $y$ and define the function $g(x)=f(x)-f(y)-x+y$. You already know $x=y$ is a solution. Suppose there is a second solution, say $x_1$. By Rolle's theorem, $\frac{dg}{dx}=0$. But, $$\frac{dg}{dx}=-\frac{x}{\sqrt{2\pi}}e^{-x^2/2}-1\neq0$$ leads to a contradiction.
