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Is it possible to affirm that if $B\subseteq \mathbb{R}$ $$f: \mathbb{R} \rightarrow B $$ and $$g: \mathbb{R} \rightarrow B$$ are invertible functions and $$f(x) \leq g(x), \forall x \in \mathbb{R} \tag 1$$ then $$g^{-1}(x) \leq f^{-1}(x) , \forall x \in B\tag 2$$ If so, how can it be proved? If it's false, is there a counterexample? In this problem, suppose $f(\mathbb{R}) = g(\mathbb{R})$.

If $f$ and $g$ are continous then this is equivalent to both are either strictly increasing or both are strictly decreasing, and the inequalities $(2)$ will hold.

Intuitively, the proposition seems true to me. I couldn't conceive a counterexample.

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    $\begingroup$ Can you please show some of your own working/thoughts on how to answer this. People are here to help you understand where you are going wrong, but not to just do questions for you. $\endgroup$ – EHH Jun 20 '17 at 14:35
  • $\begingroup$ Hint: Are $f^{-1}$ and $g^{-1}$ always defined on the same set ? $\endgroup$ – Evargalo Jun 20 '17 at 14:40
  • $\begingroup$ I unsuccessfully tried to draw a proof. Intuitively, it seems true to me. But I'm not 100% sure. So, I've tried to write the problem without say if it's true. I didn't find anything similar to this problem on Google. $\endgroup$ – cquina Jun 20 '17 at 14:47
  • $\begingroup$ @Evargalo Yes, f^{−1} and g^{−1} are defined on the same set $\endgroup$ – cquina Jun 20 '17 at 14:52
  • $\begingroup$ What if you take $f(x) = \tanh x$ and $g(x) = 4 + \tanh x$? $\endgroup$ – MPW Jun 20 '17 at 14:58
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As a hint: it is true if $f$ is increasing (in the sense that $x \le y$ iff $f(x) \le f(y)$) and $f(\mathbf{R}) \subseteq g(\mathbf{R})$. For example $f(x) = x$ and $g(x) = e^x$ then $\log x \le x \le e^x$ whenever either inequality makes sense. This isn't hard to prove.

You can use the hypotheses here to construct a counter example to what you've stated.

Something else to think about: saying that $f(x) \le g(x)$ means that $(x, f(x)$ sits below $(x, g(x)$. When you take inverses you reflect in the line $y = x$, this means that $(f(x),x)$ sits to the left of $(g(x),x)$. Why isn't this the same thing as saying that $g^{-1}(x) \le f^{-1}(x)$?


Edit (June 25, 2017):

Since this is true when $f$ is increasing, we take a decreasing $f$ for a counterexample. Let $f(x) = - x$ and let $g(x) = f(x) + 1$. Then $f(x) < g(x)$ for all $x$ but $f^{-1}(x) = f(x) < g(x) = g^{-1}(x)$ so the inequality does not reverse if $f$ is decreasing.

Claim: If $f$ is increasing then the inequality reverses.

Proof: Since $f$ is increasing, $f^{-1}$ is increasing. If $f(x) \le g(x)$ then since $f^{-1}$ is increasing,

$$ f^{-1}(f(x)) \le f^{-1}(g(x)). $$

Since $f^{-1}(f(x)) = x = g^{-1}(g(x))$ we have

$$ g^{-1}(g(x)) \le f^{-1}(g(x)) $$

or $$g^{-1}(y) \le f^{-1}(y) $$ where $y = g(x)$. Since every $y$ is of this form, the inequality holds for all $y$. $\square$

Claim: If $f$ is decreasing then the inequality does not reverse.

Proof: The same except now $f^{-1}$ is decreasing so the inequality reverses at this step:

$$ f^{-1}(f(x)) \ge f^{-1}(g(x)). $$

So we have

$$ g^{-1}(y) = f^{-1}(f(x)) \ge f^{-1}(y) $$

where $y = g(x)$. $\square$

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  • $\begingroup$ did you already construct a counter example? $\endgroup$ – miracle173 Jun 24 '17 at 9:05
  • $\begingroup$ @miracle173 I've added details/proofs. If there is anything still unclear let me know. $\endgroup$ – Trevor Gunn Jun 25 '17 at 19:28

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