# Small Challenge # 2

• Aug 1st 2010, 06:42 AM
Dinkydoe
Small Challenge # 2
This is a challenge question:

Let $f:[0,1]\to \mathbb{R}$ with $f(0)>0$ and $f(1)<0$. If there exists a continuous function $g:[0,1]\to \mathbb{R}$ such that $f+g$ is non-decreasing, show there exists a $x_0\in (0,1)$ such that $f(x_0)=0$.

Moderator edit: Approved Challenge question.
• Aug 1st 2010, 08:16 PM
Iondor
Lemma 1:
Let $\epsilon>0$ and $x \in [0,1]$.
There can be no sequence $x_{n}\uparrow x$ with $f(x)+\epsilon\leq f(x_{n})$ for all n
and
no sequence $x_{n}\downarrow x$with $f(x)-\epsilon\geq f(x_{n})$ for all n

Proof:
Towards a contradiction suppose there is such a sequence with $x_{n} \uparrow x$ .
Since $f+g$ is non-decreasing and $x_{n}\leq x$ , we have
$f(x)+\epsilon+g(x_{n}) \leq f(x_{n})+g(x_{n})\leq f(x)+g(x)$
Since g is continous, $\lim g(x_{n})=g(x)$, so that taking limits on both sides leads to the contradiction
$f(x)+\epsilon+g(x) \leq f(x)+g(x)$.
Analogous in the second case.
qed

Now consider $y:=\sup \{x\in [0,1] : f(x)\geq 0\}$. $y \in [0,1]$ , because $[0,1]$ is closed.
Assume $f(y) \neq 0$.

case 1:
$f(y)>\epsilon>0$ for some $\epsilon$.
Then $y<1$ and $f(x)<0$ for all $x>y$ according to the definition of y.
So $(y,1]$ is nonempty such that there is a sequence $x_{n} \in (y,1]$ with $x_{n}\downarrow y$ , which violates lemma 1.
case 2:
$f(y)<-\epsilon<0$ for some $\epsilon>0$
Then $y>0$.
So $[0,y)$ is nonempty such that there is a sequence $x_{n} \in [0,y)$ with $x_{n}\uparrow y$ and $f(x_n)\geq 0$ which violates lemma 1.

Therefor
$f(y)=0$.
qed
• Aug 3rd 2010, 06:55 AM
Dinkydoe
Quote:

Originally Posted by Iondor
Lemma 1:
Let $\epsilon>0$ and $x \in [0,1]$.
There can be no sequence $x_{n}\uparrow x$ with $f(x)+\epsilon\leq f(x_{n})$ for all n
and
no sequence $x_{n}\downarrow x$with $f(x)-\epsilon\geq f(x_{n})$ for all n

Proof:
Towards a contradiction suppose there is such a sequence with $x_{n} \uparrow x$ .
Since $f+g$ is non-decreasing and $x_{n}\leq x$ , we have
$f(x)+\epsilon+g(x_{n}) \leq f(x_{n})+g(x_{n})\leq f(x)+g(x)$
Since g is continous, $\lim g(x_{n})=g(x)$, so that taking limits on both sides leads to the contradiction
$f(x)+\epsilon+g(x) \leq f(x)+g(x)$.
Analogous in the second case.
qed

Now consider $y:=\sup \{x\in [0,1] : f(x)\geq 0\}$. $y \in [0,1]$ , because $[0,1]$ is closed.
Assume $f(y) \neq 0$.

case 1:
$f(y)>\epsilon>0$ for some $\epsilon$.
Then $y<1$ and $f(x)<0$ for all $x>y$ according to the definition of y.
So $(y,1]$ is nonempty such that there is a sequence $x_{n} \in (y,1]$ with $x_{n}\downarrow y$ , which violates lemma 1.
case 2:
$f(y)<-\epsilon<0$ for some $\epsilon>0$
Then $y>0$.
So $[0,y)$ is nonempty such that there is a sequence $x_{n} \in [0,y)$ with $x_{n}\uparrow y$ and $f(x_n)\geq 0$ which violates lemma 1.

Therefor
$f(y)=0$.
qed

Very good ! (Clapping)
• Aug 3rd 2010, 07:45 AM
Dinkydoe
You guys seem to have no trouble solving my Small challenges.... hehe (Rofl)

Here's what I did:

Let $h= f+g$. Then $h(0)>g(0)$ and $h(1). We need to show the existence of a $x_0\in (0,1)$ such that $h(x_0)=g(x_0)$.

Let $S = \left\{x: h(x) \geq g(x)\right\}$ and $u=\sup(S)$.

1.Suppose $h(u)>g(u)$.
Then $(u,1)$ is nonempty and by definition of $u$ we have $h(x) for all $x> u$.

Thus we must have $\lim_{x\to u^{+}}h(x)< \lim_{x\to u^{+}}g(x)=g(u)$ by continuity of $g$ . But then $h(u)>\lim_{x\to u^{+}}h(x)$. This contradicts the fact that $h$ is non-decreasing.

2. If $h(u) we get the same contradiction: $\lim_{x\to u^{-}}h(x)>\lim_{x\to u^{-}}g(x)=g(u)$. Hence $\lim_{x\to u^{-}}h(x)>h(u)$

Conclusion: $h(u)=g(u)$