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