Spring 2008 #1

Problem Statement: Let $K$ be a compact subset of $\mathbb{R}$ and $f:K\rightarrow (0,\infty)$ a continuous function. Show there exists a $\delta>0$ such that $f(x)\geq \delta$ for every $x\in K$.

Proof: Since $K$ is compact and $f:K\rightarrow (0,\infty)$ is continuous it follows that $f(K)=\{f(k)|k\in K\}$ is compact. Bolzano-Weierstrass gives us that $f(K)$ is closed and bounded. This means there is both an upper and a lower bound for $f(K)$. Since $f:K\rightarrow (0,\infty)$ is continuous it follows that there is a $x_o\in K$ such that $f(x_o)$ is a minimum since $f$ is continuous on a compact set. Let $\delta=f(x_0)$. Then $\delta$ is a lower bound for $f(K)$. We know that $\delta\neq 0$ since each $f(x)>0$ which means that the minimum value of $f$ must be strictly greater than $0$.

Thus, there exists a $\delta>0$ such that $f(x)\geq \delta$ for every $x\in K$.

Reflection: The key for this problem is that continuous functions map compact sets to compact sets. Then we know that the function attains it’s minimum value since $f$ is a continuous function on a compact set. Since we were given that $f(x)>0$ for every $x\in K$ we know that this minimum value is greater than $0$.