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

## 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

Last edited:

Dick
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.

LCKurtz
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##.

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

Last edited:
Dick
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.

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| ?

Thank you everyone for the clarification!