1. Not finding help here? Sign up for a free 30min 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!

Prove [itex]f(x)=\sqrt{x^{2}+1}[/itex] is uniformly continuous on the real line.

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove [itex]f(x)=\sqrt{x^{2}+1}[/itex] is uniformly continuous on the real line.

    2. Relevant equations

    Lipschitz Condition: If there is a constant [itex]M[/itex] such that [itex]|f(p) - f(q)| \leq M |p-q|[/itex] for all [itex]p,q \in D[/itex], then [itex]f[/itex] obeys the Lipschitz condition.

    Mean Value Theorem: Let [itex]f[/itex] be continuous on [itex][a,b][/itex] and let [itex]f'(x)[/itex] exist for all [itex]x \in (a,b)[/itex]. Then at least one point [itex]x_{o}[/itex] exists in [itex](a,b)[/itex] such that [itex]f(b) - f(a) = f'(x_{o})(b-a)[/itex].

    [itex]f'(x) = \frac{x}{\sqrt{x^{2} + 1}}[/itex] is bounded below by [itex]-1[/itex] and bounded above by [itex]1[/itex].

    3. The attempt at a solution

    Proof:
    [itex]f(x)=\sqrt{x^{2}+1}[/itex] is continuous on [itex][-N,N][/itex] and differentiable on [itex](-N,N)[/itex] for all [itex]N[/itex]. By the Mean Value Theorem, there exists an [itex]x_{o} \in (-N,N)[/itex] such that [itex]|f(x)-f(y)| \leq |f'(x_{o})||x-y|[/itex] for all [itex]x,y \in (-N,N)[/itex].

    Since the derivative [itex]f'(x_{o})[/itex] is bounded above by [itex]1[/itex] and below by [itex]-1[/itex] as [itex]x[/itex] tends to infinity, [itex]f'(x)[/itex] obeys [itex]0 < |f'(x)| < 1[/itex] for all [itex]x \in (-N,N)[/itex]. Thus [itex]0 < |f(x)-f(y)| \leq 1 \cdot |x-y|[/itex] for all [itex]x,y \in (-N,N)[/itex].

    Since these results hold for a general [itex]N[/itex], we may choose a larger [itex]N[/itex] and the results will still hold for the larger [itex]N[/itex], implying the results will hold for the real line as a whole. Letting [itex]\delta < \epsilon[/itex] thus shows that [itex]f(x)[/itex] is uniformly continuous on the real line.

    QED
     
    Last edited: Aug 22, 2012
  2. jcsd
  3. Aug 22, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've got the point, that |f'(x)| is bounded by one. So the function is Lipschitz which implies absolutely continuous. If you really want an epsilon-delta proof, go back the the definition of absolutely continuous and use the MVT from there. Working on the interval [-N,N] is not what you want for that.
     
  4. Aug 22, 2012 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your general idea is fine, but it could be written much more briefly and clearly.

    Lemma: ##|f'(x)| \le 1##
    Proof: ##|f'(x)| = \left|\frac x {\sqrt{x^2+1}}\right| < 1## is obvious.

    Proof of uniform continuity. Given ##\epsilon > 0##, let ##\delta =\epsilon##. If ##|x-y|< \delta## then by the MVT there is a ##c## between ##x## and ##y## such that ##|f(x) - f(y)| = |f'(c)(x-y)|=|f'(c)||x-y| < 1\cdot \delta = \epsilon##.

    [Edit] Dang. Dick and I tied to the minute posting answers. That takes a lot of practice. :smile:
     
    Last edited: Aug 22, 2012
  5. Aug 22, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I was trying to think of a way to express the point without using TeX. You plowed right in and did it. I'll concede priority to you. That takes time.
     
  6. Aug 23, 2012 #5
    So there's no hangup working on the whole real line from the start due to the derivative being bounded? I was unsure about using the MVT in a more general setting.
     
  7. Aug 23, 2012 #6

    Bacle2

    User Avatar
    Science Advisor


    Why would there be a problem? Like all others correctly pointed out, your Lipschitz

    constant is 1 They did all the real work, and I am just bringing up the obvious fact that

    follows from their work:

    So, if you want, say, |F(p)-F(q)|<ε , what value of δ in |p-q|<δ

    guarantees that, given that |F(p)-F(q)|<1.|p-q| ?
     
  8. Aug 23, 2012 #7
    Thank you everyone for the clarification!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook