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My proof:

I shall use the fact that the image of a closed interval under a continuous function is a closed interval (this follows from the Extreme and Intermediate Value Theorems).

Let $c,d\in (a,b)$, $c<d$. Then $f([c,d])=[m_1,M_1]$. Because $\lim_{x\to a^+}f(x)=+\infty$, $\exists \delta>0:a<x<a+\delta<c\Rightarrow f(x)>M_1$. We also have that $f([a+\delta,c])=[m_2,M_2]$. Taking $m=\min\left\{M_1,M_2\right\}$ yields that $f((a,c])=[m,+\infty)$. Similarly, $f([c,b))=[m^{\prime},+\infty)$. The proof is complete if we take the minimum of the two $m,m^{\prime}$.

Can this argument be simplified?

EDIT: We can possibly have that $a=-\infty$ or $b=+\infty$

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Local or global minimum? – yo' Dec 4 '12 at 17:54
A global minimum. – Nameless Dec 4 '12 at 17:55
up vote 2 down vote accepted

Define $g(x)=\min\{f(x), K\}$ for $x\in[a,b]$ where $K=f(x_0)$ for arbitrary $x_0$. Then show that:

  • if $g$ has a global minimum, $f$ has the same one;
  • $g$ is continuous.

Then $g$ is continuous on a closed interval $[a,b]$ therefore $g([a,b])=[c,K]$ and $c$ is the global minumum.

Proof working for $a,b$ being infinite: Let $K:=f(x_0)$ for arbitrary $x_0\in(a,b)$. From the definition of limit, there exists $A,B$ such that $f(x)>K$ for $x<A$ and $x>B$. Hence $$\min_{x\in(a,b)} f(x)=\min_{x\in[A,B]} f(x).$$

Since $f$ is continuous, the image $f([A,B])$ is closed and has a minimal element.

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what if $a=-\infty$? – Nameless Dec 4 '12 at 18:01
Then for each $K$ there exists $\tilde a\in\mathbb R$ such that $f(x)>K$ for $x<\tilde a$ (from the definition of limit). This means you can consider $a$ to be $\tilde a>-\infty$. – yo' Dec 4 '12 at 18:04
actually, you can use this argument always and completely avoid defining $g$. However, the idea is the same. – yo' Dec 4 '12 at 18:05

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