# Learning From Data – A Short Course: Exercise 8.9

Let be optimal for (8.10), and let be optimal for (8.11).

(a) Show that .

If is the optimal of (8.10) then it must satisfy the constraint , hence . If then it doesn’t matter what is, . If then we can always choose , so .

(b) Show that is feasible for (8.10). To show this, suppose to the contrary that . Show that the objective in is infinite, where as attains a finite objective of , which contradicts the optimality of .

If then solving for (8.11) is not equivalent to solving for (8.10). However, in such case will be positive infinity as we try to maximize , hence the objective in (8.11) will be infinite while obtains the finite objective in (8.10), that means is not the optimal of (8.11).

(c) Show that , and hence that is optimal for (8.10) and is optimal for (8.11).

From contradiction proof (b), we can conclude that must satisfy (the constraint in (8.10)), hence the fact (a) also follows, then what’s left in (8.11) is (the optimization problem (8.11) is now equivalent to the optimization problem (8.10)). So if is also optimal for (8.10) and is also optimal for (8.11).

(d) Let be any optimal solution for (8.11) with attained at . Show that

Either the constraint is exactly satisfied with , or .

This is already discussed in (a). Exercise 8.13 suggest that the case and may happen. I have opened a topic on this and been waiting for the answer.