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This is something I'm having a hard time finding online, but say we know that $f(x) > g(x)$ (for all inputs $x > a_{0}$ for some $a_{0}$), then would it always be true that $f^{-1}(x) < g^{-1}(x)$ (for all inputs $x > b_{0}$ for some $b_{0}$)?

For example, let $f(x) = 2^{x}$ and $g(x) = x^{2}$.

$f^{-1}(x) = \log_{2}(x)$ and $g^{-1}(x) = \sqrt{(x)}$

If I plot on desmos I see $\sqrt{(x)}$ eventually exceeds $\log_{2}(x)$, so hence we can say $f^{-1}(x) < g^{-1}(x)$ for this example

Since this is just the definition of Big-O, could we say $f = O(g) => f^{-1} = \Omega(g^{-1})$

Not sure how to prove it but isn't it intuitive from the fact that the inverse is just reflecting over $y = x$, so if $g$ is above $f$, then after reflecting $g^{-1}$ is below $f^{-1}$

RLH
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Bob Marley
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1 Answers1

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Let us look at a few simple examples to start out. $f(x) = x$ and $g(x) = x-1$. We then have $f(x)>g(x)$. We can now compute the inverses. $f^{-1}(x) =x$ and $g^{-1}(x) = x+1$ so we have $f^{-1}(x)<g^{-1}(x)$.

Now for a second example: $f(x) = -x$ and $g(x) = -x-1$. We have that $f(x) > g(x)$. we will now find the inverses $f^{-1}(x) = -x$ and $g^{-1}(x) = -x - 1$. In this case we have that $f^{-1}(x)>g^{-1}(x)$. So we can see that we cannot say for sure anything about the inverses if we know nothing about the function.

This may be too advanced, depending on your level of education, but I will say, if we expect an inverse of a continuous function to exist, then a function should be monotone otherwise an inverse would not be well defined. They should also have the same domain and range, otherwise $f(x)$ might not be defined for some $g(x)$. If these conditions are met then we can go further with your question. I claim that if $f$ and $g$ are both monotone increasing and $f(x) > g(x)$, then we have $f^{-1}(x) < g^{-1}(x)$. Additionally, if $f$ and $g$ are both monotone decreasing and $f(x) > g(x)$, then we have $f^{-1}(x) > g^{-1}(x)$. I will prove the former claim and leave the latter to you.

Let us start by assuming $f$ and $g$ are continuous, monotone increasing, and have the same domain and range. $f^{-1}$ and $g^{-1}$ are also monotone increasing. We then start with $x=x$. We then get: $$ f(f^{-1}(x)) = g(g^{-1}(x)) $$ Then because $g(y) < f(y)$, in this case $y = g^{-1}(x)$. $$ f(f^{-1}(x)) < f(g^{-1}(x)) $$ Then because $f^{-1}$ is monotone increasing: $$ f^{-1}(f(f^{-1}(x))) < f^{-1}(f(g^{-1}(x))) $$ We can then use $f^{-1}(f(y)) = y$. $$ f^{-1}(x) < g^{-1}(x) $$

chi
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Nic
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  • Thanks so much for your answer! I'm confused though about the 2 conditions you mentioned: why would the inverse not be well-defined if the the function wasn't monotone? Why does it matter for f and g to have same domain and range -- could you kindly elaborate on this part? Also, in you're claim you just assumed f & g to be monotone, not them having same domain & range. – Bob Marley May 12 '24 at 16:08
  • If they do not have the same range, then there will be a point in which $f^{-1}(x)$ is defined and $g^{-1}(x)$ is not and hence we cannot say $f^{-1}(x)<g^{-1}(x)$. Because of this I assumed, but did not state, they had the same domain and range. If they are not monotone then there will be two different points $x$ and $y$ such that $f(x) = f(y)$ and because of this the inverse will not be well defined. In your question you said the inverse of $x^2$ is $\sqrt(x)$, yet this is only true when $x>0$. So in your question, we could apply this for the domain $x\in(4,\infty)$ and the appropriate range – Nic May 12 '24 at 18:27
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    @Nic Strict monotony of (real) functions is a sufficient condition for them being bijective (i.e. having an inverse), but not a necessary one. Just as an example, consider $f(x) = 1/x$ on $\mathbb{R}\setminus{0}$ (which has itself as an inverse), but it's not monotone. – Paŭlo Ebermann May 12 '24 at 22:19
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    "if we expect an inverse to exist, then a function should be monotone" is false: @PaŭloEbermann gave an important and simple example. For another example, $f(x)=x$ if $x$ is rational and $f(x)=1-x$ if $x$ is irrational, is invertible, but there is no interval where the function is monotone.

    What "This may be too advanced" tries to convey (I guess) is that if $f$ is defined on an interval, invertible, and continuous, then $f$ has to be monotone. Opinions may vary but I think it is not helping the OP to have hidden assumptions (continuity)

     – Taladris May 12 '24 at 23:57
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    Thank you for pointing out my hidden assumption I had missed. You both are right and I edited my answer to specify that I am considering only continuous functions. If you think of a simple extension to discontinuous or a.e. continuous functions feel free to let me know and I’ll update my answer to include it. – Nic May 13 '24 at 07:40