Prove there is an absolute max on interval

Question

Suppose $f(x)$ is continuous on $(a,b)$ and $$lim_{x\to\ a^+} f(x)= lim_{x\to\ b^-} f(x) = -\infty$$

Prove that $f(x)$ has an absolute maximum on $(a,b)$.

I have no idea how I would even approach this problem. Does Rolle's Theorem have to be used?

Answer 1

Suppose $f$ doesn't have a global maximum on $(a,b)$. Then $f$ doesn't have an upper bound on $(a,b)$. Since $f$ is continous, $f$ has an upper bound on every compact interval $I(\epsilon) := [a+\epsilon, b-\epsilon]$ for every $\epsilon > 0$ (with $\epsilon$ small enough so it's still an interval). Let $x_n := \sup\limits_{x \in I\big(\frac{1}{n}\big)} f(x)$. By our assumption it follows that $x_n \to \infty$ as $n \to \infty$ and that $x_n$ is increasing. Therfore we have for

$$a_n := \sup_{x \in \big[a+\frac{1}{n+1}, a+\frac{1}{n}\big]} f(x),\ \ \ b_n := \sup_{x \in \big[b-\frac{1}{n}, b-\frac{1}{n+1}\big]} f(x)$$ that at least one of the sequences $a_n, b_n$ must diverge to $\infty$. This contradicts our assumption that $\lim_{x\to a^+} f(x) = \lim_{x\to b^-} f(x) = -\infty$

Answer 2

Take any point $x_0$ inside the interval $(a,b)$. That point has a function value $f(x_0)$.

Since $\lim_{x \to a^+}f(x) = -\infty$, there exists an $a_1 > a$ such that

$$\forall x \in (a,a_1]: f(x) < f(x_0).$$ On the other end of the interval, because of the other limit given, we can find a $b_1 < b$ with

$$\forall x \in [b_1,b): f(x) < f(x_0).$$

Obviously, $a_1 < x_0 < b_1$.

So let's now consider the continuous function $f$ on the closed and bounded interval $[a_1,b_1$]. By the extreme value theorem, it must attain a maximum $M$ on it somewhere, say $f(x_{max})=M$. Because $x_0 \in [a_1,b_1]$, we have $M \ge f(x_0)$.

But that means $f(x_{max})=M$ is the absolute maximum of $f$ even on the (larger) interval $(a,b)$, becasue on the interval $(a,a_1]$ we have

$$\forall x \in (a,a_1]: f(x) < f(x_0) \le M,$$

and similiarly for the other end of the interval

$$\forall x \in [b_1,b): f(x) < f(x_0) \le M.$$