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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?

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  • $\begingroup$ 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 $f_i(x_i)$. $\endgroup$ – The Pheromone Kid Apr 13 '15 at 9:43
  • $\begingroup$ With strongly convex I meant that the second order derivative of $f_i(x_i)$ is positive. $\endgroup$ – Thomas Apr 13 '15 at 10:45
  • $\begingroup$ Ok, I think strictly convex is term that is usally used. $\endgroup$ – The Pheromone Kid Apr 13 '15 at 10:59
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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.

I am sure there are particular cases where the two problems are equivalent, mind you. But I do not believe that it will be possible to generalize such situations too broadly.

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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 $x=0$

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