Question

Exercise 5. Prove that if f is a continuous and positive function on (0,1], there exists 8 >0 such that f(x) > 8 for any x € [0,1].

Answer 1

We will use the following lemma to prove this ,

Lemma : Continue image of a compact set is compact that is if $f:[a , b] \rightarrow \mathbb{R}$ be a continuous function then $f([a , b])$ is compact .

Given $f$ is continuous and positive function on $[0,1]$ then , $f(x)>0\, \, \, \forall \, \, \, x \in [0,1]$ .

Suppose there does not exist any $\delta >0$ such that $f(x)>\delta \, \, \forall\, x \in [0,1]$ the for all $\delta >0$​​​​​​ there exist $x \in [0,1]$ such that $f(x) < \delta$.  

Choose $\delta = \frac{1}{n}>0$ so for all $n\in\mathbb{N}$ there exist $x_{n} \in [0,1]$ such that $f(x_{n})< \frac{1}{n}$ .

So for the above choose sequence the sequence $f(x_{n})$ converges to 0 in $f([0,1])$.

Now since there exist a sequence $f(x_{n})$ converges to 0 in $f([0,1])$ so 0 a limit point of the set $f([0,1])$ and since $f([0,1])$ is compact so it contains all its limit points and so $0 \in f([0,1])$ .

$\Rightarrow$ there exist $x \in [0,1]$ such that $f(x)=0$ , a contradiction to $f(x)>0\, \, \forall \, \, x\in[0,1]$ .

Hence there exist $\delta$>0 such that $f(x)> \delta$ for any $x\in [0,1]$