Prove that a function attains its minimum

Question

Let $f$ be a real continuous function defined on $D=[0,+\infty)$, $f(x)\geq 0$ for all $x\in D$, and $\lim_{x\to+\infty}\:f(x)=+\infty$. Prove that $f(x)$ attains its minimum on $D$.

Idea for a solution

Consider the subinterval $[0,M]$, where $M>0$. By the Extreme Value Theorem, there exists $c\in[0,M]$ such that $f(c)\leq f(x)$ for all $x\in[0,M]$. Now, $f(c)=\inf f([0,M])$.

My thinking was that one could construct a sequence such that as $M$ grows large, $f(c)$ approaches the minimum of $f$. I am not sure how to do this, or if this is even the correct approach.

Answer 1

It is better to "reverse" your approach. Since $\lim_{x\to+\infty}\:f(x)=+\infty$, there exists $R>0$ such that $f(x)>f(0)$ for all $x>R$. Then, since $f$ is continuous in $[0,R]$, $$\min_{[0,R]} f(x)\leq f(0)\leq\inf_{(R,+\infty)} f(x).$$ Hence the minimum value of $f$ over $[0,+\infty)$ exists and it is equal to $\min_{[0,R]} f(x)$.

P.S. There is no need of the condition $f(x)\geq 0$ for all $x\in D$.

Answer 2

You are almost right. Since $f(x) \to \infty$ as $x\to \infty$. You can find an $M>0$ such that $f(x) > 1$ for $x>M$. So in the interval $[0,M]$ the functions $f(x) $ attains a minimum at $c$. And the minimum of $f(x) $ is $\min\{f(c), 1\} $. And you think about why there would be some $d$ such that $f(d) =1$.