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Prove functions f and g are continuous in the reals

  1. Apr 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove functions f and g are continuous in ℝ. It's known that:
    i) lim g(x)=1, when x approaches 0
    ii)g(x-y)=g(x)g(y)+f(x)f(y)
    iii)f2(x)+g2(x)=1

    3. The attempt at a solution

    g(0) has to be equal to 1 because it's known that lim g(x)=1, when x approaches 0. Otherwise g won't be continuous at x=0.

    g(0)=1 implies that f(0)=0 (iii)
    And if g(0)=1 and f(0)=0 then g(-y)=g(y) (ii) so g is even function.

    So it looks like g(x)=sin (x) and f(x)=cos(x). However, g and f may not be unique.
    f+g is continuous if f and g are continuous but not vice versa
    The definition of continuity is ∀ε>0∃δ>0: |x-x0|<δ ⇒|f(x)-f(x0|<ε ,x0∈ℝ

    How to proceed?
     
    Last edited: Apr 9, 2016
  2. jcsd
  3. Apr 9, 2016 #2

    PeroK

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    You are trying to prove that f and g are continuous, so you can't assume that g is continuous.
     
  4. Apr 9, 2016 #3
    If g(0)≠1 then g is not continuous at x=0 so g is not continuous ∀x∈ℝ and that's it. g(0)=1 is reasonable assumption.
     
  5. Apr 9, 2016 #4

    PeroK

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    You might as well go the whole way: we know (because we've been asked to prove it) that f and g are continuous, therefore they must be continuous. QED
     
  6. Apr 9, 2016 #5
    Please help me how to get started if we know nothing about g(0).
     
  7. Apr 9, 2016 #6

    PeroK

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    I would choose a point ##x_0## and write down the definition of continuity of ##g## at ##x_0##. That seems like a good place to start.

    PS I would say "to show that ##g## is continuous at ##x_0##, we need to show that ..." (not assume ##g## is continuous at ##x_0##).
     
  8. Apr 9, 2016 #7

    PeroK

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    PPS You could additionally try to find ##lim_{x \rightarrow 0} f(x)##
     
  9. Apr 9, 2016 #8
    |g(x)-g(x0)|<epsilon when |x-x0|<delta hmm

    |g(x)-g(x0)|=...?
     
  10. Apr 9, 2016 #9

    PeroK

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    That isn't a definition of anything. As an aside, if you're doing analysis you have to been more precise.

    In this case, you don't need to use epsilon-delta. You just need the properties of limits. Let me help you:

    ##g## is continuous at ##x_o## if ##lim_{x \rightarrow x_0} g(x) = g(x_0)##

    Now, I haven't given too much away, because you have to write that definition in a slightly different form. Hint: look at equation (ii) that you were given.
     
  11. Apr 9, 2016 #10
    So I won't need epsilon-delta definition at all?
     
  12. Apr 9, 2016 #11

    PeroK

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    No. Use some simple properties of limits.
     
  13. Apr 9, 2016 #12
    ##g## is continuous at ##x_o## if ##lim_{x \rightarrow x_0} g(x) = g(x_0)##⇔ ##lim_{x \rightarrow x_0} (g(x)g(0)+f(x)f(0)) = g(x_0)##
     
  14. Apr 9, 2016 #13

    PeroK

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    Let me give you one more helping hand. Then I'll be offline anyway:

    ##g## is continuous at ##x_0## if ##lim_{y \rightarrow 0} g(x_0 + y) = g(x_0)##

    Having ##y \rightarrow 0## is just another way to have ##x \rightarrow x_0##

    It might be worth looking at these two variations of the limit we have now:

    ##lim_{x \rightarrow x_0} g(x) = lim_{y \rightarrow 0} g(x_0 + y)##

    And really understand why they are equivalent.
     
  15. Apr 9, 2016 #14
    What about f?
     
    Last edited: Apr 9, 2016
  16. Apr 9, 2016 #15
    It's 0. But how to prove it?
     
  17. Apr 9, 2016 #16

    PeroK

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    Can you see how to use equation (iii)?

    Let's worry about proving ##g## is continuous first, then we can have a think about ##f##.
     
  18. Apr 9, 2016 #17
    iii)f2(x)+g2(x)=1

    ##lim_{x \rightarrow 0} g(x) =1## ⇒ ##lim_{x \rightarrow 0} g^2(x) =1##

    when x--->0 g2(x) approaches 1 so f2(x) has to approach 0 for the equation (iii) to hold. Right?

    ##lim_{x \rightarrow x_0} g(x) = lim_{h \rightarrow 0} g(x_0 - h)= lim_{h \rightarrow 0} [g(x_0)g(h)+f(x_0)f(h)]=...=g(x_0)*1+lim_{h \rightarrow 0} [f(x_0)f(h)]=g(x_0)*1+0##

    ∴ ##lim_{x \rightarrow x_0} g(x) =g(x_0)## ∀x0∈ℝ so g is always continuous
     
    Last edited: Apr 9, 2016
  19. Apr 9, 2016 #18

    PeroK

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    Yes on both counts!

    Now, you might want to wheel out the epsilons and deltas to show that:

    ##\lim_{x \rightarrow 0} f(x)^2 = 0 \ \Rightarrow \lim_{x \rightarrow 0} f(x) = 0##

    Or, you might want to just state that as "obvious" and move quickly on.

    Now, to show the continuity of ##f## follows from the continuity of ##g##.

    Hint: you said that you thought ##g(x) = cos(x)## and ##f(x) = sin(x)## but in fact ##g(x) = 1## and ##f(x) = 0## for all ##x## is also a solution.

    If ##f(x) = 0## for all ##x##, then it's continuous. Otherwise, ...
     
  20. Apr 9, 2016 #19
    if f(x)=0, then f is constant function and therefore always continuous

    if f(x)≠0 always, hmm...
     
  21. Apr 9, 2016 #20

    PeroK

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    Let me explain how I thought about it.

    First, I thought, what about equation (iii). If ##g## is continuous, then ##f^2## must be continuous. But, does ##f^2## being continuous imply ##f## is continuous? No, because ##f## could jump from positive to negative. So, perhaps I have to use equation (ii). If I picked a fixed value for ##y## then I could express ##f(x)## in terms of some combination of ##g## divided by ##f(y)## and that would do it. But, then, to do that I need ##f(y) \ne 0##. That's when I noticed that there was the constant value solution. So, ...

    If ##f## is not the zero function then ##\exists y## such that ##f(y) \ne 0## ...
     
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