1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of Odd functions' Continuity

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    If an odd function g(x) is right-continuous at x = 0, show that it is continuous at x = 0 and that g(0) = 0. Hint: Prove first that [tex]\lim_{x \to 0^{-}} g(x) [/tex] exists and equals to [tex]\lim_{x \to 0^{+}} g(-x)[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Suppose [tex]\lim_{x \to 0^{-}} g(x) = M[/tex] Let [tex]\epsilon[/tex] > 0. We must find [tex]\delta[/tex] > 0 such that whenever [tex]-\delta[/tex] < x < 0, it follows that |g(x) - M | < [tex]\epsilon[/tex]. We know that -x < [tex]\delta[/tex] but relating this to f(-x) is where I'm stuck. Like the other problem I posted, I can't see what the end result is supposed to be. Any help would be very much appreciated!
  2. jcsd
  3. Oct 19, 2007 #2
    I think your attempted solution takes it too far! For an odd function f(-x) = -f(x), that should be enough to get your proof.
  4. Oct 19, 2007 #3
    Hmm I think I do have to incorporate epsilon delta proofs in this one. I was thinking that since -g(x) = g(-x) then |-g(x)| = |g(-x)|. And since we're proving that the limit of odd continuous functions have a limit at 0 (I think that's a safe assumption) then:
    Given [tex]\epsilon>0[/tex]. [tex]\exists\delta >0[/tex] such that whenever 0 < |x| < [tex]\delta[/tex] it follows that |g(x) - M| < [tex]\epsilon[/tex]. Since M = 0, then [tex]|g(x)|<\epsilon[/tex]. (All this is the function approaching 0 from the left)

    Then looking at g(-x), for the same epsilon and delta (since the definition of odd functions implies symmetry about the origin) we arrive at the same conclusion |g(x)| < [tex]\epsilon[/tex].

    Do I seem to be on the right track or is this just dead wrong? It's difficult making the transition from concrete, repetitive epsilon-delta proofs to more general and abstract ones. The methods don't quite seem to be the same ...
    Last edited: Oct 19, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Proof of Odd functions' Continuity
  1. Continuous Function? (Replies: 5)

  2. Even or odd functions (Replies: 3)

  3. Odd or even function? (Replies: 7)