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Say I have a function of 2 or more variables f(x,y,...). There is nothing given about the function (concave, convex etc). Is it always the case that minima you get by forcing some of these variables to zero always greater than what you get otherwise?

dayum
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  • What do you mean by "forcing some of these variables to zero"? You mean, for example, given $f(x,y)$ then looking at $g(x)=f(x,0)$? Also, did you tried solving this? – ℋolo Aug 14 '18 at 23:29
  • yes exactly. I mean it is very intuitive to me, but is there a formal proof of this? – dayum Aug 14 '18 at 23:33
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    Hint, $\mbox{Range}(g)\subseteq\mbox{Range}(f)$ – ℋolo Aug 14 '18 at 23:38
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    Formal proof: fixing some variables you get a subspace, and minimum over a subset is greater than that over the whole space. – A.Γ. Aug 15 '18 at 07:08
  • Thanks @A.Γ., If you post it as answer, i will accept it. – dayum Aug 16 '18 at 01:12

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You remember me things we were doing, even with computers, about 60 years ago.

Suppose a function of two variables $f(x,y)$ that you want to minimize assuming that the minimum is unique.

  • So, fix $y=y_0$ and minimize $f(x,y_0)$ with respect to $x$; the best point corresponds to $x_0$.
  • Now, minimize $f(x_0,y)$ with respect to $y$; the best point corresponds to $y_1$. Make $y_0=y_1$ and repeat step $1$.
  • Repeat steps $1$ and $2$ until convergence.

You can generalize the procedure to any number of variables.

If I properly remember, in the very very first version of the IBM Scientific Subroutine Package, there was one subroutine performing like that

Claude Leibovici
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