# Analyze if the function is uniformly continuous: $f(x) = \frac{x}{1+x^{2}}$

Analyze if the function is uniformly continuous: $$f(x) = \frac{x}{1+x^{2}}, x\in\mathbb{R}$$

We are free to use anything but I'd like to solve it with $\varepsilon$-$\delta$. And I also think this is the best way doing, too.

I found out that the function isn't uniformly continuous but I'm not sure at all (I just claim this because I couldn't find a solution..):

$$\left|\frac{x}{1+x^{2}}-\frac{x_{0}}{1+x_{0}^{2}}\right | = \left|\frac{x(1+x_{0}^{2})-\left(x_{0}(1+x^{2})\right)}{(1+x^{2})(1+x_{0}^{2})}\right |= \left|\frac{x+x\cdot x_{0}^{2}-x_{0}-x_{0}\cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right|$$

$$= \left| \frac{x-x_{0}+x \cdot x_{0}^{2}-x_{0} \cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right| < \left|\frac{\delta + x \cdot x_{0}^{2}-x_{0} \cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right | < \left|\frac{\delta + x \cdot x_{0}^{2}}{(1+x^{2}) (1+x_{0}^{2})}\right| < \delta +x \cdot x_{0}^{2}$$

Omg, first I must say it was a pain typing all that as a LATEX beginner...

Anyway, what makes me feel bad is I cannot see if it's uniformly continuous or not, the way I solved it. Cannot even see if the function is continuous at all (but if I look at it, it seems like it's continuous at least).

Did I do anything correctly at all?

• Just take the derivative and see if it is bounded or not. Sep 20, 2016 at 18:23
• How exactly? $$f'(x) = \frac{1}{1+x^{2}} - \frac{2x^{2}}{(1+x^{2})^{2}}$$ Assuming I derivated correctly.. What now? Sep 20, 2016 at 18:27
• Check if it is bounded. You know how to do that right? Sep 20, 2016 at 18:30
• Hmm I'm not sure. Calculate limit of $f(x)$ and then set it in inequality with the derivative, then calculate $x$ and check if it's larger / smaller than the limit? Ehh no I have no idea.. :D Sep 20, 2016 at 18:34
• @tenepolis There was an error in the analysis in the OP. $$x(1+x_0^2)-x_0(1+x^2)=(x-x_0)(1-xx_0)\ne (x-x_0) x^2x_0^2$$ Sep 20, 2016 at 18:53

Note that we can write

\begin{align} \left| \frac{x}{1+x^2}-\frac{y}{1+y^2} \right|&=\left| \frac{(x-y)(1-xy)}{(1+x^2)(1+y^2)} \right|\\\\ &\le |x-y|\,\left(\frac{1+|xy|}{1+|xy|^2}\right) \end{align}

Inasmuch as $\frac{1+|xy|}{1+|xy|^2}\le \frac{1+\sqrt 2}{2}$, we find that

$$\left| \frac{x}{1+x^2}-\frac{y}{1+y^2} \right|<\epsilon$$

whenever $|x-y|<\delta =\frac{2\epsilon}{1+\sqrt 2}$. This demonstrates that the function $f(x)=\frac{x}{1+x^2}$ is uniformly continuous.

• Is there an easy way to see that $\frac{1 + |xy|}{1 + |xy|^2} \leq \frac{1 + \sqrt{2}}{2}$?
• @aduh Set the first derivative to $0$, etc. Sep 20, 2016 at 19:30
Any continuous function on $\mathbb R$ that $\to 0$ at $\pm \infty$ is uniformly continuous on $\mathbb R.$ Idea for proof: Let $\epsilon>0.$ Choose $R>1$ such that $|f(x)| < \epsilon/2$ for $|x|> R.$ Because $f$ is uniformly continuous on $[-2R,2R],$ there exists $0<\delta < 1$ such that $x,y\in [-2R,2R], |x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon.$ Suppose $y>2R$ and $|x-y|< \delta.$ Because $R>1$ and $\delta < 1,$ we then have $x,y \in [R,\infty).$ Thus $|f(x)-f(y)|\le |f(x)| + |f(y)| < \epsilon/2 + \epsilon/2 = \epsilon.$ Same for $y< -2R.$
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