# Uniform continuity of $f(x)=\arctan\frac{1}{x^2}$ on the interval $(0,\infty)$

I got this question:

Prove that the function $f(x)=\arctan\frac{1}{x^2}$ is uniformly continuous on the interval $(0,\infty)$.

I tried to prove it directly from the definition of uniform continuity but I was not able to proceed. Then I tried to prove it using the fact that $\arctan\frac{1}{x^2}=\frac{\pi}{2}-\arctan(x^2)$ and by the fact that the sum of two uniformly continuous functions is uniformly continuous but I wasn't able to prove that $\arctan(x^2)$ is uniformly continuous.

Some hints will be appreciated. Thanks.

Note: please do not use derivatives since in my class we are not allowed to use derivatives by now.

• It is true in general that a continuous, monotone, bounded function is uniformly continuous. You might consider using only these properties. – Robert Wolfe Apr 23 '14 at 16:48

I'll follow your first step and show $\arctan x^2$ is uniformly continuous on $[0,\infty)$. You can split the positive line into two parts: where $x^2$ is big, and where $x^2$ is small.

Let $\epsilon > 0$ be given.

Since $\displaystyle \lim_{x \to \infty} \arctan x^2 = \dfrac{\pi}{2}$, there exists $M > 0$ with the property that $$x > M \implies \left| \arctan x^2 - \frac{\pi}{2} \right| < \frac{\epsilon}{2}.$$ In particular, $x,y > M$ implies $|\arctan x^2 - \arctan y^2| < \epsilon$.

On the other hand, $\arctan x^2$ is continuous on $[0,M+1]$ and consequently uniformly continuous there. Thus there exists $\delta > 0$ with the property that $$x,y \in [0,M+1] \quad\text{and} \quad |x-y| < \delta \implies |\arctan x^2 - \arctan y^2| < \epsilon.$$

Let $\delta' = \min\{\delta,1\}$. Let $x,y \in [0,\infty)$ satisfy $|x-y| < \delta'$. Then either $x,y$ both belong to $(M,\infty)$ or $x,y$ both belong to $[0,M+1]$, and in either case you obtain $|\arctan x^2 - \arctan y^2| < \epsilon$.

• Thanks you very much for this answer. – MathNerd Apr 24 '14 at 4:49

Remark: In this answer we use the derivative.

The function $f$ is differentiable on $(0,\infty)$ and we have $$f'(x)=-\frac{2x}{1+x^4}$$ hence we see that $$\forall x>0,\qquad |f'(x)|\le2$$ and then by the mean value theorem $$\forall x,y>0,\qquad |f(x)-f(y)|\le 2|x-y|$$ hence $f$ is lipschitzian an then uniformly continuous on $(0,+\infty)$.

• Why the Downvote ? – derivative Apr 23 '14 at 16:43
• Excellent, Sami! – Namaste Apr 24 '14 at 12:06