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Homework Help: 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

    [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.

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


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    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


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    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


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    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


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    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!
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