For $i < j$, we know that the equation $a_i x + b_i = a_j x + b_j$ admits a solution because the affine functions are not redundant. We write $x_{i, j}$ this solution :
x > x_{i, j}&\Rightarrow& a_i x + b_i < a_j x + b_j \\
x < x_{i, j}&\Rightarrow& a_i x + b_i > a_j x + b_j
\end{array}\right.$$
For $x < x_{1, 2}$, we have $f(x) > a_2 x + b_2$. If there exists $i > 2$ such that $f(x)= a_i x + b_i$ then $x_{2, i}\leqslant x < x_{1, 2}$. So for $x' \geqslant x_{1, 2}$, we have $x' > x_{2, i}$ then $f(x)\geqslant a_i x + b_i > a_2 x + b_2$. This means that $a_2 x + b_2$ is redundant. This is a contradiction. So for $x < x_{1, 2}$, we have $f(x)= a_1 x + b_1$. Furthermore, we also have that for all $i > 2$, $x_{1, 2}\leqslant x_{2, i}$. \\
Now consider $x_{1, 2} < x < x_{2, 3}$. We know that $f(x) > a_1 x + b_1$ and $f(x) > a_3 x + b_3$. If there exists $i > 3$ such that $f(x)= a_i x + b_i$ then $x_{3, i}\leqslant x < x_{2, 3}$. So for $x' \geqslant x_{2, 3}$, we have $x' > x_{3, i}$ then $f(x)\geqslant a_i x + b_i > a_3 x + b_3$. This means that $a_3 x + b_3$ is redundant. This is a contradiction. So for $x_{1, 2} < x < x_{2, 3}$, we have $f(x)= a_2 x + b_2$. Furthermore, we also have that for all $i > 3$, $x_{2, 3}\leqslant x_{3, i}$. \\
Now consider the case $y \prec0$. We use the inequality of arithmetic and geometric means on the components of the vector $(-y_i x_i)_{1\leqslant i \leqslant n}$, which is a vector with non-negative components:
Furthermore there is equality when all $-y_i x_i$ are equal so when $x_i =-\lambda/y_i$ for $\lambda\geqslant0$. We denote by $z$ the vector such that $z_i =-1/y_i$, and we denote by $Y$ the value $\left(\prod_{i=1}^n -y_i \right)^{\frac{1}{n}}$. We then have:
Otherwise if $nY \geqslant1$ then $g(., y)$ is non-positive thanks to the inequality \eqref{geom}. Because $g(0, y)=0$ we then have $f^*(y)=0$ in this case. \\
0&\text{If } y \prec0\text{ and } n \left(\prod_{i=1}^n -y_i \right)^{\frac{1}{n}}\geqslant0\\
+\infty&\text{Otherwise}
\end{array}\right.$}
\end{center}
\paragraph{(f)}
$$ f^*(y, u)=\max_{(x, t)\in K} g(x, t, y, u)$$
$$\text{where } g(x, y)= y^\trans x + tu +\log\left( t^2- x^\trans x \right)\quad\text{ and }\quad K =\{(x, t)\in\R^n \times\R ~|~ \| x \|_2 < t \}$$
First we suppose that $\| y \|_2\geqslant-u$. We now pose $x =\lambda y$ and $t =\lambda\| y \|_2+1$ with $\lambda > 0$. We then have:
$$ g(x, t, y, u)=\lambda\| y \|_2^2+(\lambda\| y \|_2+1) u +\log\left((\lambda\| y \|_2+1)^2-\lambda^2\| y \|_2^2\right)$$
Now we use the fact that $u \geqslant-\| y \|_2$:
$$ g(x, t, y, u)=-\| y \|_2+\log\left(2\lambda\| y \|_2+1\right)\xrightarrow[\lambda\rightarrow+\infty]{}+\infty$$
So when $\| y \|_2\geqslant-u$, we have $f^*(y, u)=+\infty$. \\[2mm]
Now we suppose $\| y \|_2 < -u$. We compute the partial derivatives of $g$ with respect to $x$ and $t$:
$$\dfrac{\partial g}{\partial x}(x, t, y, u)= y -\dfrac{2x}{t^2- x^\trans x}\qquad\dfrac{\partial g}{\partial t}(x, t, y, u)= u +\dfrac{2t}{t^2- x^\trans x}$$
Let $\alpha=\frac{1}{2}\left( t^2- x^\trans x \right)$. The maxima satisfies:
$$ x =\alpha y \qquad t =-\alpha u $$
We inject this two new expressions in the derivative of $g$ with respect to $x$ and we obtain:
We check that $(x, t)$ is in $K$. We have $\| x \|_2=\alpha\| y \| < -\alpha u = t$, because we supposed $\| y \|_2 < -u$. We can now compute $g(x, t, y, u)$:
$$ g(x, t, y, u)=\alpha y^\trans y -\alpha u^2+\log\left(\alpha^2 u^2-\alpha^2 y^\trans y \right)=-2+\log\left(\dfrac{4}{u^2- y^\trans y}\right)=\log4-2-\log\left( u^2- y^\trans y \right)$$
We are finally able to give the expression of $f^*$:
