Let $ f: \mathbb{R}\longrightarrow \mathbb{R}$: compute proximal of following mapping
$$ f(x)= \sqrt {1-x^2} $$
for $ x \geq 0 $
I know that the proximal is given by
$$ \operatorname{prox}_{\!f} (x)= \operatorname{argmin}_{u\in \mathbb{R}} \big\{f(u) +(1/2)\Vert u-x\Vert^2\big\}$$
Too long to comment.
I assume that $x$ and $u$ are within the range $[-1,1]$ for $f$ to be well-defined.
Suppose $x=\sin(\theta)$, $0\leq \theta \leq \pi/2$. Let $u=\sin(\phi)$. In that case, the optimization problem is: $$ \min_{\phi}~~\cos(\phi) + \frac{1}{2}(\sin(\phi)-\sin(\theta))^2. $$ Differentiating the cost function yields: $$ -\sin(\phi) + \sin(\phi)\cos(\phi) - \sin(\theta)\cos(\phi). $$ To find the points at which the gradient vanishes, note that: $$ (\sin(\phi) - \sin(\phi)\cos(\phi))^2 = \sin^2(\theta)\cos^2(\phi). $$ Denoting $t=\cos(\phi)$, one gets the quartic polynomial: $$ t^2\sin^2(\theta) = (1-t^2)(1-t)^2. $$ This can be solved in closed form using Ferrari's method (https://en.wikipedia.org/wiki/Quartic_function), or any other numerical root finding methods. Choose the roots which lead to zero gradient (not all roots are useful), and among them the one that gives the lowest cost value.
Hope it helps.
