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Continuous inverse function

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Let I be an interval in R, and let f: I-->R be one-to-one, continuous function. Then prove that f^(-1):f(I)-->R is also continuous.

    2. Relevant equations

    3. The attempt at a solution

    I started a thread yesterday and had some responses but the proofs became quite complicated and my proof ended up wrong, so I tried more basic approach.

    So if f is continuous then
    limx->af(x) = f(a) now if f-1 is continuous then
    limf(x)->f(a)f-1(f(x)) = a
    which is equivalent to limf(x)->f(a)x=a
    which is true since f is continuous.

    Does this make any sense, I'm not really familiar with taking the limit of the domain as the function is changing. Is there any hope for this proof?

  2. jcsd
  3. Nov 4, 2008 #2
    any ideas
  4. Nov 4, 2008 #3


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    Homework Helper

    It's not a good idea to start a duplicate thread. You should've just used your previous thread. Anyway, I'll give you more detailed hints this time.

    The first order of business is to prove that f is either decreasing or increasing on I. The idea is to use the intermediate value theorem. However, since the IVT only applies to closed and bounded intervals, we need to be a little careful here. Take any two points a<b in I. Since f is one-to-one, then either f(a)<f(b) or f(a)>f(b). Suppose wlog that f(a)<f(b). Now our goal is to prove that f is increasing on I. If c<d are two other points in I, define g:[0,1]->R by

    g(t) = f((1-t)b + td) - f((1-t)a + tc).

    Why is it natural to look at this? (Draw a picture.) Use the IVT to prove that g>0 for all t. Conclude that f is increasing on I.

    Having done this, the continuity of f^(-1) follows easily from an epsilon-delta argument.
  5. Nov 4, 2008 #4
    Does the epsilon delta proof use the property f-1(f(x)) = x so

    d(x,x_o)<delta and f(x)-f(x_0)<epsilon then for the inverse

    f(x)-f(x_0)<epsilon d(f-1(f(x)) -f-1(f(x0)) )=(x-x0)<delta
    but what if delta does not equal epsilon...
  6. Nov 5, 2008 #5


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    Why are you starting with "d(x,x_o)<delta and f(x)-f(x_0)<epsilon"? We're given an epsilon and we want to find a delta>0 such that |f-1(y) - f-1(y_0)|<epsilon when |y-y_0|<delta.

    Of course everything in the domain of f-1 looks like f(something). So we may suppose that y=f(x) and y_0=f(x_0).

    Drawing a picture will make the task of finding a suitable delta easier.
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