# 1D functional equation: solve for function with given expected value w.r.t normal density

Given scalars $c_1, c_2 > 0$, how would one go about solving, for non-expansive (i.e 1-Lipschitz) $\phi: \mathbb R \rightarrow (-\infty,+\infty]$, the following equation

$$\begin{split} \mathbb E_{z \sim \mathcal N(0, 1)}[\phi(z)^2] &= c_1\\ \mathbb E_{z \sim \mathcal N(0, 1)}[\phi'(z)] &= c_2. \end{split}$$

Notation:

• $\mathcal N(0, 1)$ is the standard Gaussian normal distribution so that $$\mathbb E_{z \sim \mathcal N(0, 1)} [g(z)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp(-z^2/2)g(z)dz$$
• If the person who downvoted this post could kindly add a comment, it would be very productive. Otherwise, such attitude makes MO look like youtube, where you can downvote a random video and disappear into the wild. Thanks. – dohmatob Apr 8 '18 at 11:13

Following approach is true when $c \geq 2$.
Let $f(x) := a x$ be a linear function. Then, $\mathrm {prox}_{\lambda f}(z)= z-\lambda a$, and $$\mathbb c=E_{z \sim \mathcal N(0, 1)}[(\operatorname{prox}_{\lambda f}(z))^2] =\mathbb E_z [(z-a)^2]=2+(\lambda a)^2$$ So, $(\lambda a) ^2=c-2$, and $f (x) = (c-2)^{.5}x /\lambda^{.5}$ is a solution.