Let $x \in \mathbb{R}^n$
and, we define the following function: $f(x) = \sum_{i=1}^n (n-1-0.1i)(x_i)^2$.
This function is a convex function since it can be written as $x^T Q x $ where
$Q = \begin{pmatrix} n -1-0.1i & 0 & \ldots & 0 \\ 0 & n-1 - 0.1i & \ldots & 0 \\ \vdots & \vdots & \ddots& \vdots \\ 0 & \ldots & & n-1-0.1i \end{pmatrix} \\$
is a positive definite matrix.
I am trying to reformulate this function by using $x \in \mathbb{R}^n_+$ only. Thus, I rewrite $x_i = x_i^+ - x_i ^ -$ where $x_i^+,x_i^-\in \mathbb{R}_+$
So I write $f(x) = \sum_{i=1}^n (n-1-0.1i)(x_i^+ - x_i^-)^2$ and $Q$ becomes:
$Q^* = \begin{pmatrix} n-1-0.1 i & -(n-1-0.1i) & 0 & 0 & \ldots \\ -(n-1 -0.1i) & n-1-0.1i & 0 & 0 & \ldots \\ 0 & 0 & n-1-0.1i & -(n-1-0.1i) & \ldots \\ 0 & 0 & -(n-1-0.1i) & n-1-0.1i & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$
The new $Q^*$ is not positive definite, so $f(x)$ is not convex. How can I get this equivalent notation as strictly convex?
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).