# Thread: cts fnc on a closed interval

1. ## cts fnc on a closed interval

$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b]$ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]$

2. Originally Posted by flower3
$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b]$ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]$
Use the high-point/low-point theorem.

3. Originally Posted by flower3
$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b]$ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]$

Supose not; then for any $\displaystyle n\in\mathbb{N}\,\,\,\exists\,x_n\in [a,b]\,\,\,s.t.\,\,\,f(x_n)<\frac{1}{n}$,

Now use the continuity of the function on the sequence $\displaystyle \{x_n\}$ to reach a contradiction...

Tonio

4. Originally Posted by flower3
$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b]$ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]$
Accidental post.

5. Originally Posted by flower3
$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b]$ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]$
Alternatively to tonio's and most likely what Plato was hinting at. Since $\displaystyle [a,b]$ is closed and bounded (compact) and $\displaystyle f$ is continuous we have it that $\displaystyle f\left([a,b]\right)$ is closed and bounded (compact). Thus, $\displaystyle \inf\text{ }f\left([a,b]\right)=\alpha$ exists and is in $\displaystyle f\left([a,b]\right)$. Now since $\displaystyle \alpha\in f\left([a,b]\right)$ it follows that $\displaystyle \alpha>0$, but by the Archimedean principle we know there exists some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \frac{1}{n}<\alpha$, and the conclusion follows.