Help with rewriting a compound inequality

[Andrea94]

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).

 

[BvU]

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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[BvU]

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

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[Andrea94]

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.

 

[Andrea94]

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.

 

[BvU]

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 ...

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[Andrea94]

Right, but $\Delta g(k) - \Delta g(k-1)$ can be $\ge 0$ or $\le 0$ for an integer-convex function ... whereas ...

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Still not sure what I'm supposed to see here

 

[BvU]

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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[Andrea94]

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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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?

 

[BvU]

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 ...

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[Andrea94]

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 ...

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Ohh I see, thanks a lot for the help!