# Continuous functions/ Extreme value theorem

$g:[0,1]\rightarrow \mathbb{R}$ is continuous, $g(0)>0$, $g(1) = 0$. I want to show there is a $a\in (0,1]$ s.t. $g(a)=0$ and $g(x)>0$ for $0\le x<a$.

A student brought this to the tutoring center where I work, and the problem is in a section that comes right before the intermediate value theorem, it is in the section for extreme value theorem. So they basically have the definition of continuity and the $E.V.T$. I would love some help to find a not to intense solution. It is from a advanced calculus class.

• I don't understand how f and g are related. If you take $f=-1$ the statement seems to be false. – Felice Iandoli Nov 9 '14 at 9:58
• It should be all g's sorry – tmpys Nov 9 '14 at 22:10
• @FeliceIandoli I fixed it. – tmpys Nov 9 '14 at 22:39

Let be $x_1$ the point where $g(x_1)=\min_{[0,1]}g$. If $x_1=1$ then the exercise is solved. If $x_1\neq 1$ consider the interval $[0,\overline{x_1}]$ where $\overline{x_1}<x_1$ and $g(\overline{x_1})=0$. Let be $x_2\in [0,\overline{x_1}]$ such that $g(x_2)=\min_{[0,\overline{x_1}]}g$. Continue the construction in this way and you obtain a decreasing sequence of point $\overline{x_n}$ in $[0,1]$. Since $[0,1]$ is compact there exists $\overline x \in [0,1]$ such that $\overline{x_n}\rightarrow \overline x$. Now $\overline{x}$ is the point you were looking for because $f(\overline x)=0$ by the continuity of $f$ and $f$ is poisitive before $\overline x$ by the construction of the sequence $\overline{x_n}$.
• How do we get $g(\overline{x_1})=0$ without the intermediate value theorem? – tmpys Nov 10 '14 at 10:17