Is this proof correct? [$\lim_{x\to-\infty}f=\lim_{x\to+\infty}f=+\infty\implies\ f$ has a global minimum]

Question

I'm trying to prove that if $f : \Bbb R\to\Bbb R$ is a continuous function that verifies: $$\lim\limits_{x\to-\infty}f=\lim_{x\to +\infty}f+\infty$$

Then $f$ has a global minimum

So, since:

$\lim\limits_{x\to-\infty}f=\lim\limits_{x\to+\infty}f=+\infty\to\exists x\in\Bbb R, \exists \delta_1,\delta_2 \gt 0 \phantom{2} / \phantom{2}\forall c\in (x, x + \delta_2) :f(c)\geqslant f(x), \forall c \in (x - \delta_1, x) : f(c) \leqslant f(x) $

Since x may not be unique, letting: $$m=\min \{f(x_1),\ldots,f(x_i)\}, \phantom{2} i\in\Bbb N$$

We have that there is a global minimum at the $x_i$ of $m$

Is my reasoning correct?

Answer 1

1. It is not clear how you've used the continuity of $f$ to conclude that $f$ has any local minimum. (What you've written after the $\implies$ is precisely that $x$ is a local minimum.)
2. Even assuming that that were true, you don't know whether $f$ has finitely many local minima and so, it may not make sense to take $\min$. (In fact, you don't even know if $f$ has countably many local minima. It is actually easy to construct an $f$ which does not.)

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The correct idea would be to do something like the following:
Let $y_0 = f(0)$. Since $f(x) \to \infty$ as $x\to\pm\infty$, there exist $M_1, M_2$ with $M_1 < 0 < M_2$ such that $$f(x) > y_0 \quad \forall x < M_1$$ and $$f(x) > y_0 \quad \forall x > M_2.$$

Now, $I = [M_1, M_2]$ is compact and so, $f$ achieves its minimum on it. Let $m$ be this minimum. The claim is that this is the global minimum. Proving this is not tough. (Note that you must necessarily have $m \le y_0$ since $0 \in I$.)

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Additional note:

1. It is possible that this minimum is achieved at uncountably many points. For example, consider $f(x) = |x-1| + |x+1|$. $f$ achieves its global minimum at all points in $[-1, 1]$. However, you do have that the function values at all these points is the same.
2. It is also possible that the function has infinitely many minima of different values. For example, consider $$f(x) = \begin{cases} 0 & x = 0\\ x^2\sin\left(\dfrac1x\right) & x \in \left[-\dfrac1\pi,\dfrac1\pi\right]\setminus\{0\}\\ \left|x^2 - \dfrac{1}{\pi^2}\right| & \text{otherwise} \end{cases}$$ Here, $f$ has infinitely many distinct local minima. Thus, your original construction (which assumes that you have only finitely many mimima) will not work.

Answer 2

Let $M=\max \{f(x):|x|\le 1\},$ which exists because $f$ is continuous.

Let $r\ge 1$ such that $|y|> r\implies f(y)>M.$

Let $m=\min\{f(x): |x|\le r\},$ which exists because $f$ is continuous. And let $|x_0|\le r$ with $f(x_0)=m=\min \{f(x): |x|\le r\}$.

Clearly $|y|\le r\implies f(x_0)=m=\min \{f(x): |x|\le r\}\le f(y).$

Since $r\ge 1$ we have $$|y|>r\implies f(x_0)=\min \{f(x):|x|\le r\}\le$$ $$\le \min \{f(x):|x|\le 1\}\le$$ $$\le \max \{f(x): |x|\le 1\}=M<f(y).$$