Given the following Linear $ {L}_{1.5} $ Regression Problem:
$$ \arg \min_{x} {\left\| A x - b \right\|}_{1.5} $$
How could one redefine it as Semi Definite Programming Problem?
My (Partial) Solution
Since $ \arg \min_{x} {\left\| A x - b \right\|}_{1.5} = \arg \min_{x} {\left\| A x - b \right\|}_{1.5}^{1.5} $ one could write the problem as:
$$ \arg \min_{x} \sum_{i = 1}^{n} {\left| {a}_{i}^{T} x - {b}_{i} \right|}^{1.5} $$
Now, what I tried doing is using auxiliary variables:
\begin{alignat*}{3} % Alignment is rl \text{minimize} & \quad && {t}_{1} + {t}_{2} + \ldots + {t}_{n} & & \quad \\ \text{subject to} & \quad && {s}_{i}^{1.5} \leq {t}_{i} & & \quad i = 1, 2, \ldots, n \\ & \quad && {a}_{i}^{T} x - {b}_{i} = {s}_{i} & & \quad i = 1, 2, \ldots, n \end{alignat*}
And then form a Matrix using Schur.
Yet I can't make $ {a}_{i}^{T} x - {b}_{i} = \left| {s}_{i} \right| $.
Update
I managed to figure it out and find a tight boundary. See solution at the bottom.
One should note that for any $ p \geq 1 $ the following holds $ \arg \min_{x} \left\| f \left( x \right) \right\|_{p} = \arg \min_{x} \left\| f \left( x \right) \right\|_{p}^{p} $. Which yields:
$$ \arg \min_{x} {\left\| A x - b \right\|}_{1.5} = \arg \min_{x} {\left\| A x - b \right\|}_{1.5}^{1.5} = \arg \min_{x} \sum_{i = 1}^{n} {\left| {a}_{i} x - b \right|}^{1.5} $$
Now, one could define a tight constraints in the form of:
\begin{alignat*}{3} % Alignment is rl \text{minimize} & \quad && {t}_{1} + {t}_{2} + \ldots + {t}_{n} & & \quad \\ \text{subject to} & \quad && {s}_{i}^{1.5} \leq {t}_{i} & & \quad i = 1, 2, \ldots, n \\ & \quad && -{s}_{i} \leq {a}_{i}^{T} x - {b}_{i} \leq {s}_{i} & & \quad i = 1, 2, \ldots, n \\ & \quad && {s}_{i} \geq 0 & & \quad i = 1, 2, \ldots, n \end{alignat*}
The logic is straight forward, $ {s}_{i} $ will get as small as possible by the pressure of $ {t}_{i} $ until $ {a}_{i}^{T} x - {b}_{i} = {s}_{i} $ or $ {a}_{i}^{T} x - {b}_{i} = -{s}_{i} $ which is basically like enforcing $ \left| {a}_{i}^{T} x - {b}_{i} \right| = {s}_{i} $.
Using the appendix one could form it into SDP Form (Using $ u, v $ to from equalities from Inequalities):
\begin{alignat}{3} % Alignment is rl \text{minimize} & \quad && \boldsymbol{1}^{T} t & & \quad \\ \text{subject to} & \quad && {s}_{i} \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} + {y}_{i} \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} + 1 \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} \preceq 0 & & \quad i = 1, 2, \ldots, n \\ & \quad && {s}_{i} \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} + {y}_{i} \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} + {t}_{i} \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} \preceq 0 & & \quad i = 1, 2, \ldots, n \\ & \quad && A x + s - u = b & & \\ & \quad && A x - s + v = b & & \\ & \quad && -u, -v \preceq 0 & & \end{alignat}
Appendix
$$ {s}^{\frac{3}{2}} \leq t, s \geq 0 \Leftrightarrow \exists y : \begin{bmatrix} y & s \\ s & t \end{bmatrix} \succeq 0, \begin{bmatrix} s & y \\ y & 1 \end{bmatrix} \succeq 0 $$
This is easy to show remembering that a Symmetric Matrix $ A \in \mathbb{R}^{2 \times 2}, \; A = \begin{bmatrix} a & b \\ b & c \end{bmatrix} $ is PSD if and only if all elements on its diagonal are positive, $ a, c \geq 0 $ and $ a c \geq {b}^{2} $.