What is the set $\mathcal{D} = \{x| f(x) \leq 0\} \cap \{x| x\geq 0\}$? where $f$ is convex

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be continuous and convex. Assume that $f(0) \leq 0$. What is the set $\mathcal{D} = \{x| f(x) \leq 0\} \cap \{x| x\geq 0\}$? Assume that $\mathcal{D}$ is bounded.

I think $D = [0, \alpha]$, where $\alpha = \max\{x| f(x) = 0\}$. Any idea or proof?

Answer 1

Hint: We know that $0\in D$, since $f(0)\leq 0$. Assume that $t\in D$ with $t\neq 0$.

By definition of $D$ this means $t\geq 0$ and $f(t)\leq 0$.

Let $0<x<t$. Use convexity to show (your job) that $f(x)\leq 0$ and conclude.

Answer 2

Observe that $D$ is a closed subset of $[0,\infty)$ by the continuity of $f.$

If $D$ contains only one point, then $D=\{0\}.$ We must have $f(0)=0$ in this case. Otherwise $f(0)<0,$ which implies by continuity that $f<0$ in a neighborhood of $0.$ But then $D$ contains this neighborhood, contradicting $D$ having only one point. Thus $D=\{0\}$ and $f(0)=0,$ and the desired conclusion holds in this simple case.

In the rest of the proof I assume $D$ contains at least two points.

Suppose $a,b \in D$ with $0\le a <b.$ Claim: $[a,b]\subset D.$ This is because $f(a),f(b)\le 0,$ so the chord joining $(a,f(a))$ and $(b,f(b))$ is contained in $[a,b]\times (-\infty,0].$ The convexity of $f$ then implies $(x,f(x))$ stays below this chord for $x\in [a,b].$ Thus $f(x)\le 0$ on $[a,b],$ giving the result.

The claim shows $D$ is an interval of positive length. We also know $D$ is closed and bounded. Since $0\in D,$ we must have $D=[0,\alpha]$ for some $\alpha > 0.$ Could $f(\alpha) < 0?$ No, because if that we're true, then $f<0$ in a neighborhood of $\alpha$ by the continuity of $f.$ But then $D$ contains points greater than $\alpha,$ contradiction. So we have $D=[0,\alpha]$ and $f(\alpha) = 0,$ and this gives your conclusion.

Answer 3

First a side note: $f$ is automatically continuous, this follows from the convexity.

Define $\alpha:=\sup\{x\ |\ f(x)=0\}$. Since $D$ is bounded, we have that $\alpha<\infty$, and since $f$ is continuous, $\alpha=\max\{x\ |\ f(x)=0\}$; in particular, $f(\alpha)=0$. We now show that $\alpha\geq 0$ and $D=[0,\alpha]$.

Since $D$ is bounded, there is some $x>0$ such that $f(x)>0$. Then, since $f(0)\leq 0$, by the Intermediate Value Theorem there is some $\alpha'\in[0,x)$ such that $f(\alpha')=0$. In particular, $0\leq\alpha'\leq\alpha$.

Fix $x\in[0,\alpha]$. Then there is some $t\in[0,1]$ such that $x=t\alpha$. It follows from convexity, that $f(x)=f((1-t)0+t\alpha)\leq (1-t)f(0)+tf(\alpha)\leq 0$, since $f(0)\leq 0$ and $f(\alpha)=0$. Thus, $x\in D$, which shows $[0,\alpha]\subseteq D$.

For the converse, fix $x\in D$. Then, by definition, $x\geq0$ and $f(x)\leq 0$. Assume $x>\alpha$. Then $0\leq\alpha<x$, and there is some $t\in[0,1)$ such that $\alpha=tx$. Then, again by convexity, $0=f(\alpha)=f((1-t)0+tx)\leq(1-t)f(0)+tf(x)\leq0$, since $f(0)\leq 0$ and $f(x)\leq 0$. Thus $f(x)=0$, contradicting the definition of $\alpha$. Consequently, $x\leq\alpha$, which shows $D\subseteq[0,\alpha]$.

Answer 4

$f$ is convex so is it's sublevel (Just apply the definition of convexity of $f$) . Thus the set $\{x ~ | f(x) \leq 0 \}$ is convex (closed) subset of $\Bbb R$ so it is a closed interval. Therefore

$$D = \text{closed interval} ~ \cap ~[0 , + \infty) = [0 , \alpha] $$

Note that $\alpha < + \infty$ since we assumed $D$ is bounded.