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Is this a correct proof of a function's continuity?

  1. Nov 17, 2015 #1
    Hello,
    1. The problem statement, all variables and given/known data

    1) Let f(x) continuous for all x and (f(x)2)=1 for all x. Prove that f(x)=1 for all x or f(x)=-1 for all x.
    2) Give an example of a function f(x) s.t. (f(x)2)=1 for all x and it has both positive and negative values. Does it contradict (1) ?

    2. The attempt at a solution
    1) f(x)2 = 1 → f(x)=±1 → f(x) = 1 ∨ f(x) = -1 (since f(x) can't be both)
    2) There's no such function.
    A step function s.t.
    atrix%7D%201%20%26%20x%20%5Cgeq%203%5C%5C%20-1%20%26%20x%20%3C%203%20%5Cend%7Bmatrix%7D%5Cright..gif
    I consider incorrect since in this case f(x) =1 or f(x)=-1 not for all x but f(x)=1 for some domain and f(x)=-1 for another domain.

    Is this a correct solution ?

    Thank you
     
  2. jcsd
  3. Nov 17, 2015 #2
    No, you have to show that ##f## is always equal to 1, or always equal to -1.
     
  4. Nov 17, 2015 #3
    That's what I've shown in (1), that f(x)=1 for all x or f(x)=-1 for all x
     
  5. Nov 17, 2015 #4
    No, you only show that for any ##x##, ##f(x) \in \{-1,1\}##, which is not the same thing


    EDIT: Why is it impossible to have ##x_a,x_b\in\mathbb{R}## such that ##f(x_a) = 1## and ##f(x_b) = -1## ?
     
    Last edited: Nov 17, 2015
  6. Nov 17, 2015 #5
    ok, I see. a function can be a two-point function.
    and for (1) would it be correct to say that:
    lim f(g(x))=f(lim g(x) ):
    lim (g(x)) ^ 2 = (lim g(x) ) ^ 2 = 1 -> lim g(x) = 1 or lim g(x) = -1
     
  7. Nov 17, 2015 #6
    Why do you talk about limits? You are not looking for a local property, but a global one, aren't you ?

    EDIT: Following the hint given in post #4, what kind of set would be ##f([x_a,x_b])## and why would it be absurd given that ##1,-1 \in f([x_a,x_b])## ?
     
  8. Nov 18, 2015 #7
    Would you elaborate more on your last post, I didn't quite get you.
    Thanks
     
  9. Nov 18, 2015 #8
    The nature of ##f([x_a,x_b])## is given by the intermediate value theorem.
     
  10. Nov 18, 2015 #9

    pasmith

    User Avatar
    Homework Helper

    You contradict yourself:

    This is a function such that [itex]f(x)^2 = 1[/itex] for all [itex]x[/itex] and takes both positive and negative values. Another example is [tex]
    f(x) = \begin{cases} 1 & x \in \mathbb{Q}, \\
    -1 & x \notin \mathbb{Q}.\end{cases}[/tex] Indeed the number of such functions is equal to twice the number of ways of partitioning the reals into two disjoint non-empty subsets.

    But do any of these functions contradict the assertion that a continuous function [itex]f[/itex] satisfying [itex]f(x)^2 = 1[/itex] for all [itex]x[/itex] must be constant?
     
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