Let $I=[a,b]$ and $f:I\to \mathbb{R}$ continuous. $f(x)>0$ for $x\in I$. Prove $\exists~\alpha>0$ s.t. $f(x)>\alpha~\forall~x\in I$.

Question

This is something of a duplicate question(question), but I couldn't quite piece together the details.

My approach so far has been to try to understand the problem conceptually, then try to piece together proof from that understanding.

It stands to reason that since $f$ is continuous in $I$ then there is a maximum and minimum value for $f(x)$ where $x\in I$. (i.e. $f(c)\leq f(x)\leq f(d)$ for some $c,d\in [a,b])$

Because $c\in I$, $f(c)>0$ and using Archimidean property, there must exist some $\alpha$ s.t. $0<\alpha<f(c)$.

I'm running into trouble trying to turn this into a formal proof. Specifically, what justification do I use to declare the existence of minimum and maximum values for $f(x)$ with $x\in I$? Does the completeness property of $\mathbb{R}$ cover this?

Answer 1

Since $f$ is continuous on the closed interval $I$, by the Maximum-Minimum theorem, there exist $y\in I$such that:

$f(y)\leq f(x)$, for all $x\in I$.

We have to $f(x)>0$, for all $x\in I$, thus $f(y)> 0$, in our case define $\alpha:=f(y)$.

Thus $f(x)> \alpha$, $\forall x\in I$.

Answer 2

We shall use the Heine-Borel covering theorem. $ \newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \renewcommand\geq\geqslant \renewcommand\leq\leqslant \newcommand\bm\boldsymbol \newcommand\stpf\blacktriangleleft \newcommand\qed\blacktriangleright \newcommand\upint[2][a]{\bar {\phantom \int} \mspace{-21mu}{\int_{#1}^{#2}}} $

$\stpf$ By continuity, for each $x \in I$, $f(x) > 0$, hence there is a neighborhood $(x-\delta_x, x + \delta_x)$ that whenever $y$ is in it, $f(y) > f(x)/2$. Note that $((x-\delta_x, x+ \delta_x))_{x \in I}$ is an open covering of $I$. By the Heine-Borel covering theorem, we could extract a finite subcovering $((x_j -{\delta_j}, x_j + \delta_j))_{j=1}^N$ of $I$. Let $\alpha = \min _{j} (f(x_j)/2)$. Then $\alpha > 0$, and $\color\red{\text{for every }x \in I}$, there is some $j$ that $x \in (x_j -\delta_j, x_j + \delta_j)$, which means $\color{red}{f(x) > f(x_j)/2 \geq \alpha}$, as we desire. $\qed$