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

1. Aug 22, 2012

### nasshi

1. The problem statement, all variables and given/known data

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

2. Relevant 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 $1$.

3. 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

Last edited: Aug 22, 2012
2. Aug 22, 2012

### 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.

3. Aug 22, 2012

### 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 $|f(x) - f(y)| = |f'(c)(x-y)|=|f'(c)||x-y| < 1\cdot \delta = \epsilon$.

 Dang. Dick and I tied to the minute posting answers. That takes a lot of practice.

Last edited: Aug 22, 2012
4. Aug 22, 2012

### 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.

5. Aug 23, 2012

### 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.

6. Aug 23, 2012

### Bacle2

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| ?

7. Aug 23, 2012

### nasshi

Thank you everyone for the clarification!