Commit 64f9c4f2 authored by nanored's avatar nanored

3 subquestions left

parent 39a9de7a
......@@ -16,6 +16,7 @@
}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\trans}{\mathsf{T}}
\newcommand{\dist}{\text{\textbf{dist}}}
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We have :
$$ f^*(y) = \max_{x \in \R^n} g(x, y) \qquad \text{where } g(x, y) = y^\trans x - \max_{i = 1, ..., n} x_i $$
In the case $n = 1$ we have $g(x, y) = (y - 1) x$. Then :
\begin{center}
\fbox{$\displaystyle n = 1 \quad \Rightarrow \quad f^*(y) = \left\{ \begin{array}{lrr}
0 & \text{if} & y = 1 \\
+ \infty & \text{if} & y \neq 1
\end{array} \right.$}
\end{center} \vspace{5mm}
Now we consider $n > 1$. If there exists $i \in \{1, ..., n\}$ such that $y_i < 0$ then for $\lambda > 0$ we have :
$$ n = 1 \quad \Rightarrow \quad f^*(y) = \left\{ \begin{array}{lrr}
0 & \text{if} & y = 1 \\
+ \infty & \text{if} & y \neq 1
\end{array} \right. \vspace{5mm} $$
Now we consider $n > 1$. If there exists $i \in \{1, ..., n\}$ such that $y_i < 0$ then we let $\lambda > 0$. Because $n > 1$, we have $\max_j (-\lambda e_i)_j = 0$. Therefore we obtain :
$$ g(- \lambda e_i, y) = - \lambda y_i \xrightarrow[\lambda \rightarrow + \infty]{} +\infty $$
Where $e_i$ is the vector with a one at position $i$ and zeros everywhere else. Then if $y$ has a negative component $f^*(y) = +\infty$. \\
We now look at the case $y \in \R_+^n$. We denote by $\mathbbm{1}_n$ the vector with ones everywhere. We also denote by $m(x)$, the maximum $\max_{i = 1, ..., n} x_i$. Then $m \left( m(x) \mathbbm{1}_n \right) = m(x)$ and by positivity of $y$ we have $y^\trans x \leqslant y^\trans \left( m(x) \mathbbm{1}_n \right)$. Thus :
......@@ -217,8 +216,71 @@
\begin{center}
\fbox{$\displaystyle f^*(y) = \left\{ \begin{array}{ll}
0 & \text{if } \; y \in \R_+^n, \; \| y \|_1 = 1 \\
+ \infty & \text{else}
+ \infty & \text{otherwise}
\end{array} \right.$}
\end{center}
This is for the case $n > 1$. But we can remark that for $n = 1$ the formula is still true. So this is a correct expression in the general case.
\paragraph{(b)}
This time we have :
$$ g(x, y) = y^\trans x - \sum_{i = 1}^r x_{[i]} $$
We start with the case $r = n$. In this case we have $g(x, y) = (y - \mathbbm{1}_n)^\trans x$. So:
$$ r = n \quad \Rightarrow \quad f^*(y) = \left\{ \begin{array}{lrr}
0 & \text{if} & y = \mathbbm{1}_n \\
+ \infty & \text{if} & y \neq \mathbbm{1}_n
\end{array} \right. \vspace{5mm} $$
Now we consider the case $r < n$. As in the previous question if there exists $i$ such that $y_i < 0$, then for $\lambda > 0$, because $r < n$, we have $\sum_{j = 1}^r (-\lambda e_i)_{[j]} = 0$ and :
$$ g(- \lambda e_i, y) = - \lambda y_i \xrightarrow[\lambda \rightarrow + \infty]{} +\infty $$
Otherwise $y \succeq 0$. Suppose now that there exists $i$ such that $y_i > 1$. Then for $\lambda > 0$, we have:
$$ g(\lambda e_i, y) = \lambda (y_i - 1) \xrightarrow[\lambda \rightarrow +\infty]{} +\infty $$
We are now restricted to the case $0 \preceq y \preceq \mathbbm{1}_n$. Suppose $y^\trans \mathbbm{1}_n < r$ that is to say $\| y \|_1 < r$. Then :
$$ g(y, \lambda \mathbbm{1}_n) = \lambda (\| y \|_1 - r) \xrightarrow[\lambda \rightarrow - \infty]{} +\infty $$
In the same way if $\| y \|_1 > r$, we have :
$$ g(y, \lambda \mathbbm{1}_n) = \lambda (\| y \|_1 - r) \xrightarrow[\lambda \rightarrow + \infty]{} +\infty $$
We finally study the case where $0 \preceq y \preceq \mathbbm{1}_n$ and $\| y \|_1 = r$. We fix $x \in \R^n$. We consider the Linear Knapsack Problem (LKP) where $(x_i - x_{[n]})$'s are the (non-negative) values of $n$ objects. Where we can pick at most a quantity $1$ of each objects and where the size of our knapsack is $r$. We know that 'the' greedy algorithm is optimal for (LKP). So picking a quantity $1$ of the $r$ objects with the highest values is optimal. This means that for $y$ with $0 \preceq y \preceq \mathbbm{1}_n$ and $\| y \|_1 = r$, a vector representing the amount of each objects that we pick, we have :
$$ y^\trans (x - x_{[n]} \mathbbm{1}_n) \leqslant \sum_{i = 1}^r (x_{[i]} - x_{[n]}) $$
Because $ y^\trans \mathbbm{1}_n = \| y \|_1 = r$, we can add $x_{[n]} y^\trans \mathbbm{1}_n = r x_{[n]}$ to both sides to obtain :
$$ y^\trans x \leqslant \sum_{i = 1}^r x_{[i]} $$
So $g(x, y) \leqslant 0$ and $g(0, y) = 0$. This gives $f^*(y) = 0$ in this case. \\[2mm]
Finally we regroup all cases:
\begin{center}
\fbox{$\displaystyle f^*(y) = \left\{ \begin{array}{ll}
0 & \text{if } \; 0 \preceq y \preceq \mathbbm{1}_n, \; \| y \|_1 = r \\
+ \infty & \text{otherwise}
\end{array} \right.$}
\end{center}
This is for the case $r < n$. But we can remark that for $r = n$ the formula is still true. So this is a correct expression in the general case.
\paragraph{(c)}
\paragraph{(d)}
$$ f^*(y) = \max_{x \in \R_{++}} g(x, y) \qquad \text{where } g(x, y) = yx - x^p $$
We start with the case $p > 1$. If $y \leqslant 0$, then $g(x, y) \leqslant 0$ because $x$ is positive. Furthermore when $x$ tends to 0, $g(x, y)$ tends also to 0. So in this case $f^*(y) = 0$. \\[2mm]
Now consider $y > 0$. On $\R_{++}$, $x \mapsto x^p$ is strictly convex so $g(x, y)$ is strictly concave with respect to $x$. Then $g(., y)$ admits a unique maxima on $\R_{++}$. We derive $g$ with respect to $x$:
$$ \dfrac{\partial g}{\partial x} = y - p x^{p-1} $$
The maximum is reached when this derivative equals 0. So when $x = \left( \frac{y}{p} \right)^{\frac{1}{p-1}}$.
$$ f^*(y) \; = \; g \left( \left( \frac{y}{p} \right)^{\frac{1}{p-1}}, \, y \right) \; = \; y^{\frac{p}{p-1}} p^{-\frac{1}{p-1}} - y^{\frac{p}{p-1}} p^{-\frac{p}{p-1}} \; = \; (p - 1) \left( \frac{y}{p} \right)^{\frac{p}{p-1}} $$
We then obtain:
\begin{center}
\fbox{$\displaystyle p > 1 \quad \Rightarrow \quad f^*(y) = \left\{ \begin{array}{ll}
0 & \text{If } y \leqslant 0 \\
(p - 1) \left( \frac{y}{p} \right)^{\frac{p}{p-1}} & \text{If } y > 0
\end{array} \right.$}
\end{center}
\vspace{5mm}
Now we study the case $p < 0$. If $y > 0$ then $g(x, y)$ tends to infinity when $x$ tends to infinity. So in this case $f^*(y) = +\infty$. \\[2mm]
Then we look at the case $y \leqslant 0$. We still have the strict concavity of $g(., y)$. The derivative is still the same and the maxima has still the same expression. So we obtain the final result :
\begin{center}
\fbox{$\displaystyle p < 0 \quad \Rightarrow \quad f^*(y) = \left\{ \begin{array}{ll}
+ \infty & \text{If } y > 0 \\
(p - 1) \left( \frac{y}{p} \right)^{\frac{p}{p-1}} & \text{If } y \leqslant 0
\end{array} \right.$}
\end{center}
\paragraph{(e)}
\paragraph{(f)}
\end{document}
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