This is a question of mine that arose recently.

Let's have a real-valued function $f$ with domain $\mathbb{R}$, that is strictly decreasing and bijective on $\mathbb{R}$. Also, we have that $\lim_\limits{x\to -\infty}{f(x)}=+\infty$ and $\lim_\limits{x\to +\infty}{f(x)}=-\infty$. Prove that $\lim_\limits{x\to +\infty}{f^{-1}(x)}=-\infty$.

So, my questions are:

1) How to prove this, without infimum and supremum and without using the $\epsilon-\delta$ definition of a limit at all?

2) Do we have to have both $\lim_\limits{x\to -\infty}{f(x)}=+\infty$ and $\lim_\limits{x\to +\infty}{f(x)}=-\infty$ to conclude the result requested to prove, or is only one of those limits required?

3) Can we deduce more things about other limits and special properties of $f^{-1}(x)$, e.g. about $\lim_\limits{x\to -\infty}{f^{-1}(x)}$?

(I think that the above conclusions all seem very interesting from a mathematical aspect!)

  • 1
    $\begingroup$ You can only talk about $f^{-1}$ when $f$ is a bijection, with the current conditions, this is not true. $\endgroup$ Mar 29 '16 at 17:51
  • 1
    $\begingroup$ Do you assume $f$ to be continuous? $\endgroup$
    – Crostul
    Mar 29 '16 at 17:52
  • $\begingroup$ strictly monotonic functions are invertible? $\endgroup$
    – user90533
    Mar 29 '16 at 17:53
  • 3
    $\begingroup$ @Jason If $f$ is not continuous, then $f^{-1}$ does not exists. $\endgroup$
    – Hetebrij
    Mar 29 '16 at 17:54
  • 1
    $\begingroup$ @Hetebrij: That's not entirely true. Rather, if $f$ is strictly monotonic but not continuous, then $f$ is invertible, but the domain of its inverse is not all of $\Bbb R.$ $\endgroup$ Mar 29 '16 at 17:56

From the comments, I use the following "definition" for $\lim_{x\to \infty} f(x) = -\infty$: $$ \forall K \in \mathbb R : \exists x\in\mathbb R : f(x) < K. $$ In general it is not the same, but for decreasing functions it is. $\lim_{x\to -\infty} f(x) = \infty$ is "defined" analogously.

There is nothing fancy here.

  1. Notice that $f^{-1}$ is strictly decreasing as for $y_1,y_2\in f(\mathbb R)$ with $y_1 = f(x_1) > f(x_2) = y_2$ we have $x_1 < x_2$, otherwise it would contradict that $f$ is strictly decreasing.

  2. Let $K\in\mathbb R$. Then, from $\lim_{x\to -\infty} f(x) = \infty$ there exists some $x\in\mathbb R$ with $f(x) > f(K)$, thus $f^{-1}(y) < f^{-1}(f(K)) = K$ for $y=f(x)$. That is $\lim_{y\to \infty} f^{-1}(y) = -\infty$.

    On the other hand from $\lim_{x\to \infty} f(x) = -\infty$ there exists some $\tilde x\in\mathbb R$ with $f(\tilde x) < f(K)$, thus $f^{-1}(\tilde y) > f^{-1}(f(K)) = K$ for $\tilde y = f(\tilde x)$. That is, $\lim_{y\to-\infty} f^{-1}(y) = \infty$.

  • $\begingroup$ May you, if you have the kindness, provide a similar proof using the $\epsilon-\delta$ austere limit definition? Just underneath the current proof. Also, thank you very much!! $\endgroup$
    – user171110
    Mar 29 '16 at 21:42
  • $\begingroup$ @Jason: It is basically the same. The $\epsilon$-$\delta$ definition for $\lim_{x\to\infty} f(x) = -\infty$ would be $\forall K\in\mathbb R : \exists x_0\in \mathbb R : \forall x > x_0 : f(x) < K$. $\endgroup$
    – user251257
    Mar 29 '16 at 21:44

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