Commit d464dec9 authored by nanored's avatar nanored

Fin DM 1

parent 64f9c4f2
......@@ -253,6 +253,40 @@
\paragraph{(c)}
$$ f^*(y) = \max_{x \in \R} g(x, y) \qquad \text{where } g(x, y) = yx - max_{i = 1, ..., m} \; (a_ix + b_i) $$
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 :
$$ x_{i, j} = \dfrac{b_j - b_i}{a_i - a_j} \qquad \text{and } \left\{ \begin{array}{lll}
x > x_{i, j} & \Rightarrow & a_i x + b_i < a_j x + b_j \\
x < x_{i, j} & \Rightarrow & a_i x + b_i > a_j x + b_j
\end{array} \right.$$
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 $i > 3$, $x_{2, 3} \leqslant x_{3, i}$. \\
By recurrence we end up obtaining :
$$ x_{1, 2} \leqslant x_{2, 3} \leqslant ... \leqslant x_{m-1, m} \qquad
f(x) = \left\{ \begin{array}{llc}
a_1 x + b_1 & \text{if} & x < x_{1, 2} \\
a_i x + b_i & \text{if} & x_{i-1, i} \leqslant x < x_{i, i+1} \\
a_m x + b_m & \text{if} & x_{m-1, m} \leqslant x
\end{array} \right. $$
Then for $y < a_1$ and $x < x_{1, 2}$, we have:
$$ g(x, y) = x (y - a_1) - b_1 \xrightarrow[x \rightarrow -\infty]{} + \infty $$
For $y > a_m$ and $x > x_{m-1, m}$, we have:
$$ g(x, y) = x (y - a_m) - b_m \xrightarrow[x \rightarrow +\infty]{} + \infty $$
Finally for $a_i \leqslant y \leqslant a_{i+1}$ we compute the derivative of $g$ with respect to $x$:
$$ \dfrac{\partial g}{\partial x}(x, y) = \left\{ \begin{array}{llc}
y - a_1 & \text{if} & x < x_{1, 2} \\
y - a_j & \text{if} & x_{j-1, j} \leqslant x < x_{j, j+1} \\
y - a_m & \text{if} & x_{m-1, m} \leqslant x
\end{array} \right. $$
Then $g(., y)$ is non-decreasing for $x \leqslant x_{i, i+1}$ and non-increasing for $x \geqslant x_{i, i+1}$. Then:
$$ f^*(y) = g(x_{i, i+1}, y) = (y - a_i) \dfrac{b_j - b_i}{a_i - a_j} - b_i $$
We can now write:
\begin{center}
\fbox{$\displaystyle f^*(y) = \left\{ \begin{array}{llc}
+\infty & \text{if} & y < a_1 \text{ or } a_m < y \\
(y - a_i) \dfrac{b_j - b_i}{a_i - a_j} - b_i & \text{if} & a_i \leqslant y \leqslant a_{i+1}
\end{array} \right.$}
\end{center}
\paragraph{(d)}
$$ f^*(y) = \max_{x \in \R_{++}} g(x, y) \qquad \text{where } g(x, y) = yx - x^p $$
......@@ -280,7 +314,56 @@
\end{center}
\paragraph{(e)}
$$ f^*(y) = \max_{x \in \R_{++}^n} g(x, y) \qquad \text{where } g(x, y) = y^\trans x + \left( \prod_{i=1}^n x_i \right)^{1/n} $$
First we can remark that when there exists $i$ such that $y_i > 0$ then :
$$ g(\lambda e_i, y) \geqslant \lambda y_i \xrightarrow[\lambda \rightarrow +\infty]{} +\infty $$
So in this case, $f^*(y) = +\infty$. \\[2mm]
In the same way, if there exists $i$ such that $y_i = 0$ then for $\lambda > 0$ we have:
$$ g(\lambda e_i + \mathbbm{1}_n, y) = y^\trans \mathbbm{1}_n + (\lambda + 1) \xrightarrow[\lambda \rightarrow +\infty]{} +\infty $$
So we have again, $f^*(y) = +\infty$. \\[2mm]
Now consider the case $y \prec 0$. 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:
$$ - \dfrac{y^\trans x}{n} \geqslant \left( \prod_{i=1}^n (-y_i x_i) \right)^{\frac{1}{n}} = \left( \prod_{i=1}^n -y_i \right)^{\frac{1}{n}} \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} $$
We then obtain:
\begin{equation}
\left( 1 - n \left( \prod_{i=1}^n -y_i \right)^{\frac{1}{n}} \right) \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \geqslant g(x, y)
\label{geom}
\end{equation}
Furthermore there is equality when all $-y_i x_i$ are equal so when $x_i = -\lambda/y_i$ for $\lambda \geqslant 0$. 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:
$$ (1 - nY) \lambda / Y = g(\lambda z, y) $$
If $nY < 1$ then:
$$ g(\lambda z, y) \xrightarrow[\lambda \rightarrow +\infty]{} +\infty $$
Otherwise if $nY \geqslant 1$ 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. \\
Finally we obtain:
\begin{center}
\fbox{$\displaystyle f^*(y) = \left\{ \begin{array}{ll}
0 & \text{If } y \prec 0 \text{ and } n \left( \prod_{i=1}^n -y_i \right)^{\frac{1}{n}} \geqslant 0 \\
+ \infty & \text{Otherwise}
\end{array} \right.$}
\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:
$$ y = \dfrac{2 \alpha y}{\alpha^2 u^2 - \alpha^2 y^\trans y} = \frac{1}{\alpha} \dfrac{2y}{u^2 - y^\trans y} \quad \Rightarrow \quad \alpha = \dfrac{2}{u^2 - y^\trans y} \geqslant 0 $$
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 $g(x, t, y, u)$:
$$ g(x, t, y, u) = \alpha y^\trans y - \alpha u^2 + \log \left( \alpha^2 u^2 - \alpha^2 y^\trans y \right) = -2 + \log \left( \dfrac{4}{u^2 - y^\trans y} \right) = \log 4 - 2 - \log \left( u^2 - y^\trans y \right) $$
We are finally able to give the expression of $f^*$:
\begin{center}
\fbox{$\displaystyle f^*(y, u) = \left\{ \begin{array}{ll}
+ \infty & \text{If } \| y \|_2 \geqslant -u \\
\log 4 - 2 - \log \left( u^2 - y^\trans y \right) & \text{If } \| y \|_2 < -u
\end{array} \right.$}
\end{center}
\end{document}
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