Help with rewriting a compound inequality

  • #1
Andrea94
21
8
TL;DR Summary
Help with rewriting optimality conditions for integer-convex functions.
See attached screenshot.
Stumped on this, I'll take anything at this point (hints, solution, etc).

help.png
 

Answers and Replies

  • #2
BvU
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Big problems are to be split into smaller problems. You have two inequalities. Write the first one down and manipulate ! Then idem number two.

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  • #3
BvU
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And: what does integer-convex mean for e.g. $$g(k+1) + g(k-1) - 2g(k) \quad \textsf{?} $$

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  • #4
Andrea94
21
8
Big problems are to be split into smaller problems. You have two inequalities. Write the first one down and manipulate ! Then idem number two.

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Great, thanks! Didn't even think about the fact that I could do each inequality separately.
 
  • #5
Andrea94
21
8
And: what does integer-convex mean for e.g. $$g(k+1) + g(k-1) - 2g(k) \quad \textsf{?} $$

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What do you mean? The expression you've written is the same thing as $$\Delta g(k) - \Delta g(k-1)$$ but I'm not sure how that is relevant.
 
  • #6
BvU
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What do you mean? The expression you've written is the same thing as $$\Delta g(k) - \Delta g(k-1)$$ but I'm not sure how that is relevant.
Right, but ##\Delta g(k) - \Delta g(k-1)## can be ##\ge 0## or ##\le 0## for an integer-convex function ... whereas ... :wink:

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  • #7
Andrea94
21
8
Right, but ##\Delta g(k) - \Delta g(k-1)## can be ##\ge 0## or ##\le 0## for an integer-convex function ... whereas ... :wink:

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Still not sure what I'm supposed to see here 😅
 
  • #8
BvU
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The problem with inequalities is that you can only multiply (or divide) left and right with something positive. As it happens, for integer-convex functions the sign of ##g(k+1) + g(k-1) - 2g(k)## is ...

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  • #9
Andrea94
21
8
The problem with inequalities is that you can only multiply (or divide) left and right with something positive. As it happens, for integer-convex functions the sign of ##g(k+1) + g(k-1) - 2g(k)## is ...

##\ ##
Hm, the only thing I can think of is that if ##k## is optimal and nonzero, then ##g(k+1) + g(k-1) - 2g(k)## is always positive since for optimal ##k## we have ##\Delta g(k) > 0## and ##\Delta g(k-1) < 0##. Is this what you mean?
 
  • #10
BvU
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Not sure where your 'optimal' comes from (it seems to live in your context, but not in the context of this thread ?).

But: yes, for a convex function the second derivative is always positive and so is this ##g(k+1) + g(k-1) - 2g(k)##.

I figured it might help in manipulating the inequalities ...

##\ ##
 
  • #11
Andrea94
21
8
Not sure where your 'optimal' comes from (it seems to live in your context, but not in the context of this thread ?).

But: yes, for a convex function the second derivative is always positive and so is this ##g(k+1) + g(k-1) - 2g(k)##.

I figured it might help in manipulating the inequalities ...

##\ ##
Ohh I see, thanks a lot for the help!
 

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