I have the following convex optimisation problem which is assumed to give an optimal solution:
$max: f(x)$
$ a_{i}\leq x \leq b_{i}$ for $i=1,...,n$ and $a_{i}< b_{i}$ where $a_{i}, b_{i} \in \mathbb{R}^{n}$
$f$ is convex and differentiable over $\mathbb{R}^{n}$
I am trying to show that the KKT conditions hold by showing that strong duality holds and admits a dual solution using slaters condition. However, seeing as these are box constraints I am struggling to choose a sufficient feasible solution $x^{*}$ such that $f_{i}(x^{*})<0$ where $f_{i}(x)$ are the inequality constraints define above.
I have tried to re-write the constraints as $x_{i} - b_{i}\leq 0$ and $a_{i} - x_{i}\leq 0$ and look for an $x^{*}$ related to $b_{i} - a_{i} <0$ which we know to satisfy slater's conditions.
Any suggestions?
Any help is greatly appreciated
Slater's condition only "applies" to nonlinear functions. Or put another way, it always holds for linear constraints. To be a Slater point, it only need be in the strict interior relative to nonlinear constraints. That is sufficient for strong duality.
As for KKT, the Linear Constraint Qualification applies. Therefore, KKT is necessary for this problem, presuming that $f(x)$ is continuously differentiable.
Note, I am assuming you made a "typo", and that $f(x)$ is concave (not convex), which is needed to make the maximization problem convex.