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*The full question is: *
If $f:R→R$ is a continuous increasing bijection function and $g:R→R$ is a continuous decreasing bijection function, then prove or disprove that $h(x)=f(x)−g(x)$ is a bijection function.

Am not good at analysis, know of the concept of continuity (from left & right, based on limits) only. This concept helps in deriving properties needed for my basic knowledge.
If two functions are monotonic (added to emphasize further the increasing/decreasing property), bijective (implies to me- that range exists for entire domain, but significance is not intuitive in this context) continuous functions, then even if the domain (& also range) of one is a subset (need not be proper), then still $h(x)$ 'will' not be bijective, as shown below:

If $f$ is suppose increasing, and $g$ decreasing over the same interval (both are having same domain and ranges, with none subset of the other, and are equal); then can imagine (with no value assigned to domain, function, hence to range to substantiate) that one function ($f$) is increasing with same domain and range, while another ($g$) is decreasing over the same domain (with the same range). Hence, the difference is zero, which forms the range of $h(x)$ and is a constant value, and hence is not a bijection function even, as $h(x)$ is having the same value for all elements. In case, the range elements have a constant difference for $f, g$, then still the same situation remains.

José Carlos Santos
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jiten
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  • Do you know the intermediate value theorem? – Angina Seng Jun 02 '18 at 17:57
  • @LordSharktheUnknown It is easy & intuitive too (both graphically, & algebraically), and states : if a continuous function $f$, with an interval $[a, b]$ as its domain, takes values $f(a)$ and $f(b)$ at each end of the interval, then it also takes any value between $f(a)$ and $f(b)$ at some point within the interval. – jiten Jun 02 '18 at 18:00
  • @LordSharktheUnknown Couldn't link IVT to this problem, please elborate. – jiten Jun 02 '18 at 19:16
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    JCS has already done so. – Angina Seng Jun 02 '18 at 19:19

2 Answers2

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The functions $f$ and $-g$ are both increasing and bijective. Therefore;

  • $\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}-g(x)=+\infty$;
  • $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}-g(x)=-\infty$.

So, $f-g$ is injective (since it is strictly increasing) and surjective ($\lim_{x\to\pm\infty}f(x)-g(x)=\pm\infty$ and surjectivity follws then from the intermediate value theorem).

José Carlos Santos
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Guide:

  • To prove that a fucntion is a bijection function, we have to check that they are both injection and surjection.

  • To prove that it is an injection, suppose that $x_1 \ne x_2, h(x_1)=h(x_2)$, then we obtained $$f(x_1)-g(x_1)=f(x_2)-g(x_2)$$

$$f(x_1)-f(x_2)=g(x_1)-g(x_2)$$

Try to see a contradiction.

  • To show surjection, note that $\lim_{x \to -\infty} f(x)-g(x)=- \infty$ and $\lim_{x \to \infty} f(x)-g(x)=\infty$, note that $f-g$ is also a continuous function and we can use intermediate value theorem on $h$ to show surjection.
Siong Thye Goh
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  • Please see my post at :https://math.stackexchange.com/q/2806115/424260, as am confused over the different approaches shown. My approach is based on logic, and request vindication; else reason why am wrong. – jiten Jun 03 '18 at 04:31
  • Please detail either here or on the post (let, a) (at: https://math.stackexchange.com/q/2804677/424260) that caused this OP (let, b), the reason how this fits into that post (a). The reason for confusion is that in post (a), the two functions have been proved to have only one unique point in common as common domain. So, the single point is the only common domain. – jiten Jun 03 '18 at 04:39
  • sorry, no idea what you are talking about. – Siong Thye Goh Jun 03 '18 at 04:44
  • I mean that in the earlier post (at : https://math.stackexchange.com/q/2804677/424260) there is only one unique point in common between the two functions (as given in the comment to that post by @ArnaudMortier- that stated two functions- one bijective increasing given by $x+a^3$, another bijection decreasing given by $\sqrt[3]{a-x}$). I hoped that your question : to prove/disprove .... (as comment to that post) was due to the same reasons. Was confused that with 1 unique point as common between 2 domains, your question (as comment) was significant. If so, why? Also request a chat in that post. – jiten Jun 03 '18 at 05:04
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    I was trying to give you a justification of why the solution exists and unique. Let $f(x)=x+a^3$ which is continuous increasing bijection and $g(x)=\sqrt[3]{a-x}$ which is continuous decreasing bijection. Hence $f(x)=g(x)$ has a unique solution since the question is equivalent to $f(x)-g(x)=0$ which is equivalent to $h(x)=0$. As for finding the solution explicitly, the answer that was given was elegant. – Siong Thye Goh Jun 03 '18 at 05:09
  • Have two questions, both very imp. to me. First, if only one point is there in the common domain, still there is need to prove the difference (i.e., $h(x)$) is also a bijection . The only possibility averse to it could be if domain had more than one points mapping to the same point of range (unable to find that range element-pls. help). But if domain has only 1 point (where $g(x), f(x)$ match), then why need check bijection. 2nd, why is solution elegant. I have seen 6 votes & am unable to find reason why all are calling it elegant. It pains me as if am mathematically illiterate. – jiten Jun 03 '18 at 05:31
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    You just need to find a solution and show that it is an injection. I am just giving you a more general result. As for why it is elegant, for me a single substitution that leads to a symmetrical equation $a+a^3=t+t^3$ is cool. – Siong Thye Goh Jun 03 '18 at 05:40