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

[nasshi]

Homework Statement

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

Homework Equations

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

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

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

The Attempt at a Solution

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

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

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

QED

 

[Dick]

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.

 

[LCKurtz]

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.

 

[Dick]

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.

 

[nasshi]

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.

 

[Bacle2]

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.

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| ?

 

[nasshi]

Thank you everyone for the clarification!