# Small Challenge # 2

• Aug 1st 2010, 05:42 AM
Dinkydoe
Small Challenge # 2
This is a challenge question:

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

Moderator edit: Approved Challenge question.
• Aug 1st 2010, 07:16 PM
Iondor
Lemma 1:
Let $\displaystyle \epsilon>0$ and $\displaystyle x \in [0,1]$.
There can be no sequence $\displaystyle x_{n}\uparrow x$ with $\displaystyle f(x)+\epsilon\leq f(x_{n})$ for all n
and
no sequence $\displaystyle x_{n}\downarrow x$with $\displaystyle f(x)-\epsilon\geq f(x_{n})$ for all n

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

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

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

Therefor
$\displaystyle f(y)=0$.
qed
• Aug 3rd 2010, 05:55 AM
Dinkydoe
Quote:

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

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

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

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

Therefor
$\displaystyle f(y)=0$.
qed

Very good ! (Clapping)
• Aug 3rd 2010, 06:45 AM
Dinkydoe
You guys seem to have no trouble solving my Small challenges.... hehe (Rofl)

Here's what I did:

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

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

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

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

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

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