The problem is given by:
$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \mu {\left\| x \right\|}_{1} \\
\text{subject to} \quad & {\left\| x \right\|}_{2} \leq 1
\end{aligned}$$
Which is equivalent to
$$\begin{aligned}
\arg \min_{x} \quad & \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \mu {\left\| x \right\|}_{1} \\
\text{subject to} \quad & {\left\| x \right\|}_{2}^{2} \leq 1
\end{aligned}$$
The Lagrangian is given by:
$$ L \left( x, \lambda \right) = \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \mu {\left\| x \right\|}_{1} + \lambda \left( {x}^{T} x - 1 \right) $$
With similar reasoning to the solution of Orthogonal Projection onto the $ {L}_{2} $ Unit Ball and derivation of the Proximal Operator of the $ {L}_{1} $ Norm one could conclude that (By examining the case $ \lambda \neq 0 $ and the case $ \lambda = 0 $):
- If the norm of $ y $ after the soft thresholding is larger than 1 then one should apply the projection onto the $ {L}_{2} $ Unit Ball.
- If the norm of $ y $ before or after the soft thresholding is smaller or equal to 1 then then one should apply only the Soft Thresholding.
Hence the solution is given by:
$$ x = \frac{ \operatorname{sign} \left( y \right ) {\left( \left| y \right| - \mu \right)}_{+} }{ \max \left( 1, \operatorname{sign} \left( y \right ) {\left( \left| y \right| - \mu \right)}_{+} \right) } = \frac{ \mathcal{S}_{\mu} \left( y \right) }{ \max \left( 1, \mathcal{S}_{\mu} \left( y \right) \right) } $$
Where $ \mathcal{S}_{\mu} \left( \cdot \right) $ is the Soft Threshold operator with parameter $ \mu $ and $ {\left( \cdot \right)}_{+} $ denotes the positive part.
To verify the result I implemented it in MATALB and verified it against CVX.
The code is available at my StackExchange Mathematics Q2057347 GitHub Repository.
