How are linear programs always convex?

Question

In Convex Optimization by Boyd, it is stated that

Since any linear program is therefore a convex optimization problem, we can consider convex optimization to be a generalization of linear programming.

An optimization problem is called linear, if the objective and constraint functions are all linear, i.e. satisfy
$f_i(\alpha x + \beta y) = \alpha f_i(x) + \beta f_i(y)$ for all $x, y \in \mathbb{R}^n$, and all $\alpha, \beta \in \mathbb{R}$.

A convex optimization is one where the objective and constraint functions are convex, they satisfy
$f_i(\alpha x + \beta y) \leq \alpha f_i(x) + \beta f_i(y)$ for all $x, y \in \mathbb{R}^n$, and all $\alpha, \beta \in \mathbb{R}$ with $\alpha + \beta = 1, \alpha \geq 0, \beta \geq 0$.

How are linear programs having $\alpha, \beta$ not in this range say $4, 6$ respectively also convex?

Answer 1

Convex optimization problem states that we just have to check the condition for $\alpha, \beta$ nonnegative and sums to $1$.

For linear function, we have to check all the real $\alpha, \beta$. Since it holds for any $(\alpha, \beta) \in \mathbb{R}^2$, it will hold when there is a restriction as well.

Answer 2

linearity 's attributes as follow

• $f(ax) = af(x)$
• $f(x+y) = f(x)+f(y)$

if $f(ax+by) = af(x)+bf(y) $ we can divide by a+b in both side, that is $f(ax+by)/(a+b) = f(ax/(a+b)+by(a+b))$(attribute 2)$ = af(x)/(a+b)+bf(y)/(a+b)$ which is satisfied with definition of convex optimization.