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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

    micromass

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    This looks ok to me!!

    And indeed, you don't need that absolute value in the end because the function is monotone.
     
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