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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.

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  • $\begingroup$ Do you know the intermediate value theorem? $\endgroup$ – Lord Shark the Unknown Jun 2 '18 at 17:57
  • $\begingroup$ @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. $\endgroup$ – jiten Jun 2 '18 at 18:00
  • $\begingroup$ @LordSharktheUnknown Couldn't link IVT to this problem, please elborate. $\endgroup$ – jiten Jun 2 '18 at 19:16
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    $\begingroup$ JCS has already done so. $\endgroup$ – Lord Shark the Unknown Jun 2 '18 at 19:19
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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.
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  • $\begingroup$ Please see my post at :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. $\endgroup$ – jiten Jun 3 '18 at 4:31
  • $\begingroup$ Please detail either here or on the post (let, a) (at: 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. $\endgroup$ – jiten Jun 3 '18 at 4:39
  • $\begingroup$ sorry, no idea what you are talking about. $\endgroup$ – Siong Thye Goh Jun 3 '18 at 4:44
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    $\begingroup$ 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. $\endgroup$ – Siong Thye Goh Jun 3 '18 at 5:09
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    $\begingroup$ 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. $\endgroup$ – Siong Thye Goh Jun 3 '18 at 5:40
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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).

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  • $\begingroup$ Could not understand why limit of difference of two functions is not a constant, or zero; as in my post. $\endgroup$ – jiten Jun 2 '18 at 18:05
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    $\begingroup$ @jiten SInce $f$ and $-g$ are increasing, their sume (that it, $f-g$) is increasing too. $\endgroup$ – José Carlos Santos Jun 2 '18 at 18:06

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