# Convex optimization problem: linear equality and inequality constraints

When linear equality constraints can be converted in an inequality constraints for a strongly convex optimization problem? I mean, I got the same solution for both the following problem:

1)

$\min_x \sum_i f_i(x_i)$

s.t $Ax=b$

2)

$\min_x \sum_i f_i(x_i)$

s.t $Ax\geq b$

where $f_i$ are strongly convex for each $i$ and $x$ is the vector of $x_i$.

There exists a formal proof for explaining the result?

• What do you mean by stronly convex? In general, the structure of your functions $f_i(x_i)$ can lead to equivalent solutions for 1) and 2). Therefore, it is helpful to further define $f_i(x_i)$. – The Pheromone Kid Apr 13 '15 at 9:43
• With strongly convex I meant that the second order derivative of $f_i(x_i)$ is positive. – Thomas Apr 13 '15 at 10:45
• Ok, I think strictly convex is term that is usally used. – The Pheromone Kid Apr 13 '15 at 10:59

There is no formal proof, because it is not always true. In fact, it is possible to select $A$ and $b$ such that (2) is feasible but (1) is not. The properties of the objective function are basically irrelevant. You just happen to be lucky for your particular instance.
No. Take $f(x)=x^2$ and $A=1$, $b=-1$ and then you will see that the optimal solution for first problem is $x=-1$ but for second one is $x=0$