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Consider the following function $$\pi_{p}(x) = \left(\frac{1}{2}x_{1}-x_{2}\right)^{2}+\left(\frac{1}{4}x_{1}^{2}-\rho \min(0,x_{1}-1)\right)$$

For any $\rho$ we want to minimize this function. The minimum of $\pi$ is obtained at a point where $x_2 = \frac{1}{2}x_{1}$ and where $x_1$ minimizes the function defined by $$\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),& \text{if} \ x_1 <1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} \geq 1 \end{cases}$$

Shouldn't it be $$\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),&\text{if} \ x_1 \geq 1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} < 1 \end{cases}$$

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If $x_1<1$, then $x_1-1<0$, so $\min(0,x_1-1)=x_1-1$, and you want $\frac14x_1^2-\rho(x_1-1)$. If $x_1\ge 1$, then $x_1-1\ge 0$, so $\min(0,x_1-1)=0$, and you want simply $\frac14x_1^2$.

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