Minimum of a function on an open interval

Question

Suppose that f is continuous on $(a,b)$ and $\lim_{x\to a^+} f(x) = \lim_{x\to b^-} f(x) = \infty$. Prove that $f$ has a minimum on all of $(a,b)$.

I was thinking of applying the extreme value theorem, but I realized that it applies only to closed intervals. So I was thinking of defining a closed interval that is in $(a,b)$, and using the definition of one sided limits to go about the problem, but I don't know how to proceed. Help please?

Answer 1

Take a sequence of nested compact intervals contained in $(a,b)$ that 'converge' to $(a,b)$. What can you say about the minimum of $f$ in each of these compact intervals? What can you say about the values $f$ assumes outside these compact intervals?

Answer 2

The conditions over the left and right limits on $a$ and $b$ imply that for some $n\geq1$ there exists $x_0\in[a+1/n,b-1/n]$ such that $$f(x_0)<f(x)\text{ for every }x\in(a,a+1/n)\cup(b-1/n,b).\hspace{1cm}(\ast)$$

This allows us to restrict our attention to $[a+1/n,b-1/n]$, where things get easier, for $(\ast)$ implies that $$\inf_{x\in(a,b)}f(x)=\inf_{x\in[a+1/n,b-1/n]}f(x),$$ and for $[a+1/n,b-1/n]$ is compact (recall that a continuous function in a compact set always attains its minimum).

Answer 3

Take $x_0\in(a,b)$. Because the limits in $a$ and $b$ are $+\infty$, we have $f(x)>f(x_0)$ on some neighborhoods of $a$ and $b$, namely $(a,a+\epsilon_1)$ and $(b-\epsilon_2,b)$. The minimum is not attained in those intervals.

We can now apply the minimum value theorem on $[a+\epsilon_1, b-\epsilon_2]$ and find the minimum for all $(a,b)$

BONUS: if you want to be fancy you might also consider $g:(a,b)\rightarrow (-\pi/2,\pi/2)$ defined by $g(x)=\tan^{-1}\circ f(x)$ which has a continuous extension to $[a,b]$ by defining $g(a)=g(b)=\pi/2$. Apply the minimum value theorem for $g$ and the minimum found $x_0\in(a,b)$ is also a minimum for $f$.

Answer 4

The function has an "intermediate value" property (https://en.wikipedia.org/wiki/Intermediate_value_theorem). You cannot use it directly (as in $(a,b)$ you have no $a$ and $b$), but this is more or less a reason of the fact you prove.

Let us take a sequence of sub-intervals $[a_i,b_i]$ s.t. $a_i\to a,\ b_i\to b$.

On each $[a_i, b_i] \ f$ has a minimum $m_i$. Now it is enough to show that from some pint on the minimum does not change.

Suppose towards contradiction that it is changing all the time. The only possibility for it to change is when $f$ has lower and lower values close to $a$ or $b$ (or both) - otherwise it would be captured by the previous intervals. However, both $f(a_i)$ and $f(b_i)$ diverge to positive infinity, so it is not possible.