I am stumped on this problem

Question

It is from Intro to Analysis by Bartle 3rd ed Let $I:=[a,b]$ let $f:I\rightarrow\mathbb{R}$ be a continuous function with $f(x)>0$ Prove that there exists a number $c>0$ such that $f(x)\geq c $ for all in $I$

I am given 2 hints: Use Boundedness theorem with $1/f$ or Max-min theorem

If $f$ is bounded then $1/f$ is unbounded? Can this type of theorem be done without proof by contradiction? Mock proof If $f$ is continuous then it is bound. So we have $|f(x)|<M$ For any number $n$ in $\mathbb{N}$ there is a number $x_n$ in $I$ s.t $|f(x_n)|<n$ Ugh help

Answer 1

$f$ is a continuous function on a compact set, so its image is compact (this is probably the content of the “min-max” theorem in your book); hence $f(I)$ is of the form $[c,d]$ so that for all $x\in I$, $f(x)\ge c>0$. That $c>0$ follows from the fact that $0\not\in f(I)$.

I should also remark that $f$ bounded does not imply $1/f$ unbounded: take $f(x) \equiv 1$ defined on $[0,1]$ for instance.

Answer 2

Without compactenss (directly, at least. This is Calculus I stuff): use Weierstrass theorem II: the function attains its maximum and minimum on $\;[0,1]\;$, and since $\;f(x)>c\;$ for all $\;x\in[0,1]\;$ , the minimum must be then positive.

Answer 3

Proof using first hint: $\frac 1 f$ is a continuous function on I, hence it is bounded. Suppose $\frac 1 {f(x)} <M$ for all $x$. Then $f(x) >\frac 1 M$ for all $x$. Take $c=\frac 1 M$.

Proof using second hint: there exists a point $x_0$ such that $f(x_0)$ is the minimum value of $f$ on $I$. Take $c=f(x_0)$ and note that $c=f(x_0) \leq f(x)$ for all $x$.