Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$

Question

Let $x,y \in \mathbb{R} $ such that;

If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$

$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1 \leq y^2 +1 -1 < 2\delta := \epsilon $$

Now, this is the part I am having trouble with if $|x| > \delta$

$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| \leq |\frac{x^2 - y^2}{\sqrt{x^2 +1 } + \sqrt{y^2 +1 }}| \color{red}{\leq \frac{(x+y)^2}{2\sqrt{\delta^2 + 1}}}...$$

I don't think the red term makes much sense. As the $2xy$ term could very well be negative.

Could someone suggest a next step please.

Answer 1

You came close. We have $$\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|=\frac{|x^2-y^2|}{\sqrt{1+x^2}+\sqrt{1+y^2}} =|x-y|\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}. $$ Now $$\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\le \frac{|x|+|y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\lt 1. $$

Answer 2

Usually, before trying to attack this kind of problems with the definition of uniform continuity, it is much easier and much faster to try to prove that the functions involved are Lipschitz-continuous.

In your case,

$$\left| \sqrt{x^2 + 1} - \sqrt {y^2 + 1} \right| = \left| \frac {x^2 - y^2} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| = \left| \frac {x + y} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| \cdot |x - y| = \left| \frac {x + y} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| \cdot |x - y| \le \left( \frac {|x| + |y|} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right) \cdot |x - y| \le 1 \cdot |x-y| ,$$

so $\sqrt {x^2 + 1}$ is Lipschitz-continuous of Lipschitz constant $1$, whence uniform continuity follows immediately (remember that one characterization of uniform continuity is: $f$ is uniformly continuous on $I$ if and only if for $|x_n - y_n| \to 0$, we have $|f(x_n) - f(y_n)| \to 0$ for $(x_n)_n, (y_n)_n \subset I$).