Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this inverse function continuity proof consistent?

  1. Nov 22, 2011 #1
    I am self-studying Calculus and tried to solve the following question:

    1. The problem statement, all variables and given/known data
    Suppose that the function f is continuous and increasing in the closed interval [a, b]. Then
    (i) f has an inverse f-1, which is defined in [f(a), f(b)];
    (ii) f-1 is increasing in [f(a), f(b)];
    (iii) f-1 is continuous in [f(a), f(b)].
    The book does the proofs for (i) and (ii). It then proofs (iii) partially by showing that f is continuous in the open interval (a,b).
    I must then show that f-1 is continuous in f(a) to the right and continuous in f(b) to the left.

    2. Relevant equations
    Definition of unilateral limit to the right:
    [tex]\lim_{x\to a^+} f(x) = L[/tex]
    if and only if, for every small ε > 0, there is a δ > 0 such that
    |f(x) - L| < ε whenever 0 < x - a < δ

    3. The attempt at a solution
    I will try to do here the proof that f-1 is continuous in f(a) to the right.
    Applying the definition of unilateral limit to the right, I must then show that, for every small ε > 0, there is a δ > 0 such that
    |f-1(y) - a| < ε whenever 0 < y - f(a) < δ

    I begin by defining ε such that a + ε belongs to [f(a), f(b)].
    Since a < a + ε, then, since f is increasing:
    [tex]f(a) < f(a+\epsilon)[/tex]
    Now, since the choice of δ depends upon the choice of ε, I choose δ such that:
    [tex]\delta \leq f(a+\epsilon) - f(a)[/tex]
    [tex]f(a)+\delta \leq f(a+\epsilon)[/tex]
    Now, from the definition of the limit above, I know that:
    [tex]0 < y - f(a) < \delta[/tex]
    Summing f(a) to every term:
    [tex]f(a) < y < f(a) + \delta[/tex]
    So, combining this with the δ chosen above:
    [tex]f(a) < y < f(a) + \delta \leq f(a+\epsilon)[/tex]
    Since f and f-1 are both increasing:
    [tex]f^{-1}(f(a)) < f^{-1}(y) < f^{-1}(f(a+\epsilon))[/tex]
    [tex]a < f^{-1}(y) < a+\epsilon[/tex]
    Subtracting a from every term:
    [itex]0 < f^{-1}(y) - a < \epsilon[/itex] whenever [itex]0 < y - f(a) < \delta[/itex]
    which appears to be the desired result, except for the lack of absolute value. I imagine the absolute value didn't appear here because [itex]f^{-1}[/itex] is an increasing function.

    Thank you in advance.
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2
    This looks ok to me!!

    And indeed, you don't need that absolute value in the end because the function is monotone.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook