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The continuity of f

  1. Feb 6, 2013 #1
    I have seen this theorem in a few books, but none of them give proofs, it says

    if f(x) is a continuous function then lf(x)l is a continuous function. What is the proof of this because i don't really understand why this holds, thanks
  2. jcsd
  3. Feb 6, 2013 #2
    The function [itex]|~|:\mathbb{R}\rightarrow \mathbb{R}[/itex] is continuous. Composition of continuous functions is continuous.
  4. Feb 6, 2013 #3


    Staff: Mentor

    It makes sense though.

    Consider y=-1/x : its continuous in the domain of x in (0,infinity) right?

    and its counterpart z=|-1/x| = 1/x is continuous in the same domain for x
  5. Feb 6, 2013 #4
    Ahh, yeah i get that, but can you not prove it without using composition of two functions?
  6. Feb 6, 2013 #5


    Staff: Mentor

    You could probably prove it via the Weierstrass definition:


    but Im not a mathematician so you'll have to hope HallsOfIvy checks this thread.
  7. Feb 6, 2013 #6
    i might pm him, cheers
  8. Feb 6, 2013 #7
    It is very easy to prove using the [itex]\epsilon-\delta[/itex] definition. Are you familiar with [itex]\epsilon[/itex]and [itex]\delta[/itex] definitions?
  9. Feb 6, 2013 #8
    yeah, that's about the limit of my analysis knowledge, how would you use the ε,δ definition?
  10. Feb 6, 2013 #9


    Staff: Mentor

    No need micro can handle it too. I think the epsilon / delta is the weierstrass definition.
  11. Feb 6, 2013 #10
    Take an [itex]a\in \mathbb{R}[/itex]. You will need to prove

    [tex]\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow ||f(x)|-|f(a)||<\varepsilon[/tex]

    You are given that f is continuous in a, thus:

    [tex]\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon[/tex]

    So, take [itex]\varepsilon>0[/itex] arbitrary. Take [itex]\delta>0[/itex] as in the previous definition: so it holds that

    [tex]\forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon[/tex]

    Take an x arbitrary such that [itex]|x-a|<\delta[/itex]. Then we know that

    [tex]||f(x)|-|f(a)||\leq |f(x)-f(a)|<\varepsilon[/tex]

    So we have verified that [itex]||f(x)|-|f(a)||<\varepsilon[/itex] and thus we have verified that [itex]\varepsilon-\delta[/itex] definition of continuity. Thus [itex]|f|[/itex] is continuous.
  12. Feb 6, 2013 #11
    ah that makes sense and is very obvious, do you need to show that limit of lf(x)l is in fact lf(a)l or is that just obvious?
  13. Feb 6, 2013 #12
    I've just proven that here. It's because |f| is continuous in a.
  14. Feb 6, 2013 #13
    oh okay, i get it, thanks for the help
  15. Feb 8, 2013 #14
    Are you trying to prove this for a class or so that you understand it? Judging from your posts, I don't think that you understand the epsilon-delta statements.

    Many calculus teachers give the intuitive rule "if you can completely draw the graph of f on an interval without lifting your pencil, then f is continuous on that interval." This is true, and it will work if you are given an f for which the antecedent is true. (In your case, you're not given a particular f, so this rule would not lead to a proof, even on an informal level.)

    However, mathematicians do not care about continuous functions because they appear to be connected; they study continuous functions because continuous functions enjoy the property that small changes in input do not significantly affect their outputs. That is, if we want our function outputs to be within some margin of error ε about the function value f(y), we can always bound an interval of radius δ about y so that if we choose any input in the interval (y-δ,y+δ), we are guaranteed to have an output within the interval (f(y)-ε,f(y)+ε). Any function that enjoys this property is continuous at y.

    Sorry if I'm telling you something you already know, but I have had far too many students who still equate continuity with connectivity because teachers of the calculus never tell the students the true importance of continuous functions. I thought I'd better intervene before you settled on what continuity means in your mind.
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