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For a convex function $h(x)$, what conditions must hold for $$g_2(x) = ah(x) - b, \\ a, b \in R$$ is also convex?

My intuition is that for a convex $h(x)$ then $a > 0$ because if $a < 0$ then we 'flip' the function and it becomes concave. I also think that $b \in R$ because this will only shift the function vertically. How can I formalise this in a mathematical manner?

InvestingScientist
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1 Answers1

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Here's a hint:

All convex functions $f : A \rightarrow B$ satisfy, for $t \in [0,1]$ and $x\leqslant y \in A$:

$$f((1-t)x+ty) \leqslant (1-t)f(x) + tf(y)$$

Can you verify that $g_2$ is convex given that $h$ satisfies this condition?

Your intuition that $a \geqslant 0$ is something you should listen to.

While I Am
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  • Ok, so I think I have got it. Using the definiton of convexity as you proposed, multiplication by an scalar $a \geq 0$ leaves the inequality unchanged. For $a < 0$ we reverse the inequality leading to convexity. Then if we consider the case where we subtract a constant from both sides, this does not change the sign of the inequality and thus our sufficient conditions are that $a \ geq 0$. Have I missed anything? – InvestingScientist Mar 18 '21 at 20:59
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    Sounds like a good proof template. You should be careful, though--- subtracting $b$ from both sides and scaling both sides by $a$ is technically correct, but is sort of skipping steps. Instead, try replacing $f(x)$ with $af(x) - b$ wherever $f$ appears. Also, can you see how it's both necessary and sufficient? – While I Am Mar 18 '21 at 23:54