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Suppose $x$ is $n$-dimensional real, and $f(x)$ and $g(x)$ are real and convex R1.

I want to find $x$ to minimize $Q= f(x)+g(x)$ s.t. $a’x=b$, for $a$ $N \times 1$ real vector and constant b. Let's call this solution $X^*$. How is $X^*$ related to the two partial solutions $X_1$ and $X_2$ that optimize the following?

  • $X_1$ is the solution to $\min Q_1 = f(x)$ s.t. $a’x=b$.

  • $X_2$ is the solution to $\min Q_2 = g(x)$ s.t. $a’x=b$.

If no direct/simple connection, can i use $X_1$ and $X_2$ in an efficient manner in an interative optimization scheme, such as ADMM or augmented Lagrangian, to arrive at $X^*$ ?

MH-NY
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  • Welcome to mathSE. If you take all the equations in your post and enclose them in two dollar signs, they will display nicely and be easier to read! And X1 can be done as $X_1$, so it makes $X_1$. – Caleb Stanford Oct 01 '16 at 20:10
  • Thank u. Will learn and improve. – MH-NY Oct 01 '16 at 20:13
  • I edited your question to show you! Please click "edit" and you can see what I did. – Caleb Stanford Oct 01 '16 at 20:21
  • Thank u so much. I really appreciate ur help. – MH-NY Oct 01 '16 at 20:27
  • Yes, saw the edits. Will try next time. Once again, thank u. – MH-NY Oct 01 '16 at 20:29
  • Honestly, I don't see a whole lot that you can do with $X_1$ and $X_2$, save perhaps using some convex combination of those two points as the initial point in an iterative method (e.g., the midpoint). You're simply solving a different problem when those two objectives are added. – Michael Grant Oct 02 '16 at 18:50
  • The initial condition that seems interesting is to choose $X_0$ = c$X_1$+(1-c)$X_2$, and choose c to minimize Q, but I con't show if this initial condition is better/worse than any other. – MH-NY Oct 03 '16 at 10:33

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