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I had a homework problem which asks to prove or disprove that A is a convex set where $A=\{x: g(x) \le c\}$. and $g(x)$ is a convex function. I went ahead in this way:

Assume $x_1$ and $x_2$ $\in$ A.

$g(x_1)\le c$ $\rightarrow$ $g^{-1}(g(x_1)) \le g^{-1}(c)$ $\rightarrow$ $x_1=g^{-1}(c)-\alpha$ where $\alpha \ge 0$

$g(x_2)\le c$ $\rightarrow$ $g^{-1}(g(x_2)) \le g^{-1}(c)$ $\rightarrow$ $x_2=g^{-1}(c)-\beta$ where $\beta \ge 0$

$tx_{1}+(1-t)x_2$ = $tg^{-1}(c)+(1-t)g^{-1}(c)-t\alpha-(1-t)\beta$=$g^{-1}(c)-t\alpha-(1-t)\beta$ $\le g^{-1}(c)$

as $\alpha, \beta, t, (1-t) \ge 0$

$tx_{1}+(1-t)x_2\le g^{-1}(c)\rightarrow g(tx_{1}+(1-t)x_2)\le g(g^{-1}(c)) \rightarrow g(tx_{1}+(1-t)x_2)\le c$

which proves the set is convex. However, this does not use the information that g is convex, which seems weird. As you can see, I am assuming the inverse functions exist, can I assume that? Or the proof is wrong altogether?

Asaf Karagila
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rivu
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2 Answers2

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Assume $x,y$ are in $A$. We can say $g(x) \le c$ and $g(y) \le c$. Now we have to investigate weather $g(\gamma x + (1-\gamma)y) \le c$ or not? Since $g(x)$ is convex we can say (convex function property)

$$g(\gamma x + (1-\gamma)y) \le \gamma g(x) + (1-\gamma)g(y)$$

and since $x,y \in A$ we have $\gamma g(x) + (1-\gamma)g(y) \le \gamma c + (1-\gamma)c = c$. Therefore we concluded that $\gamma x + (1-\gamma)y \in A$ thus $A$ is a convex set.

K.K.McDonald
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I'm assuming you mean a single variable function $g(x)$. If $c \leq \min_{\forall x} g(x)$, then $A$ is the empty set or one-set, which is degenerate and is convex.

If $c > \min_{\forall x} g(x)$, then $g$ intersects exactly twice with the line $y = c$, and therefore, $A = \{x: x\in [a,b]\}$, where $a,b \in \mathbb{R}$, which can be verified to be a convex set.

laughing_man
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