I would like to prove this
Let $G(y)\doteq\mathbb{E}[X|\gamma X+\gamma^{1/2}Z=y]$ where $Z\sim\mathcal{N }(0,1)$ and $X\sim\pi(x)$ where $\pi(x)$ has bounded support. $Z$ and $X$ are indepenedent. I want to prove that $G(y)$ is $||G||_{\infty}^2$-Lipschitz and non-decreasing.
This is my attempt. I denote $p_Y(y)$ the density of $Y=\gamma X+\gamma^{1/2}Z$. Since $X$ and $Z$ are indip. $$p_Y(y)=\frac{1}{\gamma^{3/2}} \pi(\cdot/\gamma)*\phi(\cdot/\gamma^{1/2})$$
I want to use Tweedie's formula $$\mathbb{E}[X|Y=y]=y+\gamma\partial_y\log p_Y(y)$$
So I have a more explicit expression for the function $G(y)=y+\gamma\partial_y\log p_Y(y)=y+\gamma\frac{1}{p_Y(y)}\partial_yp_Y(y)$
From here I've tried to take the derivative of $G$ and I hoped that is non negative and bounded by $||G||_\infty^2$ so that I can have the thesis, so $$\partial_yG(y)=1+\gamma[-(\frac{1}{p_y(y)})^2(\partial_yp_Y(y))^2+\frac{1}{p_Y(y)}\partial_{yy}p_Y(y)]=1+\gamma[-(\frac{G(y)-y}{\gamma})^2+\frac{1}{p_Y(y)}\partial_{yy}p_Y(y)]$$
Now I write explicitly$$p_Y(y)=\frac{1}{\gamma^{3/2}}\int\pi(x/\gamma)\exp[-(y-x)^2/\gamma]dx$$This is the convolution of $\pi(x/\gamma)\in L^1$ and $\exp(-x^2/\gamma)\in C^\infty\cap L^\infty$ and its derivatives belong to $L^\infty$ as well. So $p_Y(y)\in C^\infty$ and its derivative is calculated simply "taking the derivative inside the integral" so$$\partial_{yy}p_Y(y)=\frac{1}{\gamma^{3/2}}\int\pi(x/\gamma)\partial_{yy}\exp[-(y-x)^2/\gamma]dx=\frac{1}{\gamma^{3/2}}\int\pi(x/\gamma)\exp[-(y-x)^2/\gamma]\frac{4(y-x)^2-2\gamma}{\gamma^2}dx$$
In conclusion I would like to prove that$$1+\gamma[-(\frac{G(y)-y}{\gamma})^2+\frac{1}{p_Y(y)}\partial_{yy}p_Y(y)]$$is bounded by $||G||_{\infty}^2$ and non negative, but I cannot conclude? Am I on the right path?