4

If a function $f$ from $\mathbb{R}$ to $\mathbb{R}$ is one-to-one and bounded is it true that $f^{-1}$ is also one-to-one and bounded? I believe the answer is no but I'm not sure.

covertbob
  • 433
  • The inverse is one-to-one (otherwise, $f$ wouldn't be a function). – David Mitra Jan 14 '13 at 03:39
  • Technically, a bounded function from R to R can't have inverse – Max Jan 14 '13 at 09:40
  • I think that $ f^{-1} $ here is defined as $ f^{-1} \stackrel{\text{def}}{=} { (b,a) \in \mathbb{R}^{2} ~|~ (a,b) \in f } $. Therefore, $ \text{Dom}(f^{-1}) = \text{Range}(f) $ and $ \text{Range}(f^{-1}) = \text{Dom}(f) $. – Haskell Curry Jan 16 '13 at 05:56

4 Answers4

10

The answer is always false. The range of $f^{-1}$ is the domain of $f$, which from the way the problem is stated is $\mathbb R$.

N. S.
  • 132,525
5

A counter example: Logistic function.

4

No. Consider $\arctan x$ which is bounded, but $\tan \theta$ isn't bounded.

Calvin Lin
  • 68,864
  • +1. Interesting to note that both of us had similar kind of sketch of the function in our mind. –  Jan 14 '13 at 03:33
  • @Marvis Seems the easiest to use. I wanted a 'simple' example, though OP should have seen Logistic functions under exponentials too. Just wondering, are you an analyst? Your 'calculus/analysis' proofs are very beautiful. – Calvin Lin Jan 14 '13 at 04:34
  • Thanks and the answer to your question is "I don't know" :-). I am a doctoral student in computational mathematics working in numerical linear algebra/ algorithms. I just dabble in different areas of math with courses I have taken. –  Jan 14 '13 at 04:49
  • And brilliant.org looks like a great initiative! Good one. Is there a way I can contribute to it? –  Jan 14 '13 at 07:45
  • @Marvis Thanks! Certainly, we're constantly looking for good problems for students to work on, in which they can demonstrate their problem solving ability. We haven't expanded into calculus/analysis as yet, though we plan to do so in the near future :) – Calvin Lin Jan 18 '13 at 00:49
  • 1
    Sure. I am also interested in Olympiad/Putnam type problem solving in general. Anyway, in case you guys plan to expand and need someone to contribute some time down the line, do keep me posted. :-) –  Jan 18 '13 at 00:52
  • I am interested in opportunities at brilliant. I saw through MathJobs the position of Mathematics Challenge Master but the deadline (2012/11/02) is over. Are you still accepting applications? If so and if you do not mind me emailing you, can you give me your email id. My id is here. Thanks. –  Jan 25 '13 at 22:43
  • @Marvis Done. Was about to email you actually :) – Calvin Lin Jan 25 '13 at 22:50
0

Inverse function itself may not exist. For example the function $f(x) = \frac{1}{1+e^x}$ is defined from $\mathbb{R}$ to $\mathbb{R}$ which is one to one and bounded. But its inverse $\log\left(\frac{1}{y} -1\right)$ is not defined for all real numbers.

Stefan Hansen
  • 25,582
  • 7
  • 59
  • 91