To derive the proximal operator of $f=\frac{1}{2} \Vert x \Vert_2^2$:
$\mathbb{prox}_{\lambda f}(x)=\left(\frac{1}{1 + \lambda} \right) x$
following, what is the proximal operator of the squared error $f=\frac{1}{2} \Vert x - x^{true} \Vert_2^2$ ?
The solution that I found is:
$\mathbb{prox}_{\lambda f}(x)=\left(\frac{1}{1 + \lambda} \right) \left(x + x^{True} \right)$
How did they drive that solution ?
Let $H$ be a Hilbert space, $c \in H$, and $f:H \to \mathbb R$ be the function defined by $f(x) := \|x-c\|^2/2$. Given $a \in H$, in order to compute $\text{prox}_{\lambda f}(a)$, you must solve the problem
$$ \text{Minimize } \frac{1}{2}\|x-a\|^2+\lambda f(x) \text{ over }x \in H. $$
The KKT conditions are: $a-x \in \lambda\partial f(x) = \{\lambda (x-c)\}$, from which we get $$ x=(a+c)/(1+\lambda), $$ as claimed.
N.B.: $\partial f(x) := \{v \in H \mid f(z) \ge f(x)+\langle v,z-x\rangle\,\forall z \in X\}$ denotes the subdifferential of $f$ at $x$.