This post is the follow-up to Uniform Contnuity of Real-Valued Functions. Now let \(f\) be our real-valued function \(\frac{1}{x^2+1} \) defined on the set of real numbers \(\mathbb{R}\). Our first proof that \(f\) is uniformly continuous on \(\mathbb{R}\) really just requires a few inequality manipulations.

Let \(\epsilon >0\), \(\delta=\frac{\epsilon}{2}\), and \(x,y \in \mathbb{R}\). We then have for whenever \(x,y \in \mathbb{R}\) and \(|x-y| < \delta \) that \( |f(x)-f(y)|=\left|\frac{1}{x^2+1}-\frac{1}{y^2+1}\right|=\left|\frac{y^2-x^2}{(x^2+1)(y^2+1)}\right|=|x-y| \frac{|x+y|}{(x^2+1)(y^2+1)} \leq |x-y| \left(\frac{|x|}{(x^2+1)(y^2+1)}+\frac{|y|}{(x^2+1)(y^2+1)}\right) \).

At this point it is important to observe that for any real number \(w >0\), \(w < w^2+1\) and hence \(\frac{w}{w^2+1} < 1\). To prove that \(w < w^2+1\) for all \(w > 0\), just break it into two cases where one case is when \(0<w<1\) and the other when \(w \geq 1 \).

It now follows that because \( \frac{1}{(x^2+1)}, \frac{1}{(y^2+1)}, \frac{|y|}{(y^2+1)}, \frac{|x|}{(x^2+1)} < 1\), we have that \( \left(\frac{|x|}{(x^2+1)(y^2+1)}+\frac{|y|}{(x^2+1)(y^2+1)} \right) \leq 2 \) and hence \( |x-y| \left(\frac{|x|}{(x^2+1)(y^2+1)}+\frac{|y|}{(x^2+1)(y^2+1)}\right) \leq |x-y|\cdot 2 < \epsilon\)

Another proof only requires applying the the Mean Value Theorem, the derivative to \(f\), and observing then after that \(f\) is Lipschitz. Here is this proof:

First notice that \(f\) is continuous and differentiable on \(\mathbb{R}\). Applying the Mean Value Theorem we have that for any two elements \(x > y \in \mathbb{R}\), \( \frac{f(x)-f(y)}{x-y}=f'(c)\) for some \(c \in (x,y)\) and hence \( |f(x)-f(y)|=|f'(c)||x-y|\). We also have that \(f'(x)=\frac{-2x}{(x^2+1)^2}\) and so \(|f'(x)|\) is bounded from noticing that \(\left| \frac{-2x}{(x^2+1)^2} \right|= 2\frac{|x|}{(x^2+1)^2}<2\). Now we're ready to finish this proof.

Let \(\epsilon >0\) and \(\delta=\frac{\epsilon}{2}\). We then have that \(|f(x)-f(y)|=\left|\frac{1}{x^2+1}-\frac{1}{y^2+1} \right|=|f'(c)||x-y| < 2|x-y|<\epsilon \). We have thus shown that for \(\epsilon >0\), there exists a \(\delta >0\) such that if \(|x-y|<\delta\), then \(|f(x)-f(y)|<\epsilon\). An alternative is to say that because \(f'\) is bounded by some bound \(M\), we have that \(f\) is Lipschitz on \(\mathbb{R}\) since \(|f(x)-f(y)|<M|x-y|\). Either way, we therefore have that \(\frac{1}{x^2+1} \) defined on the set of real numbers is uniformly continuous.

Before reading the last proof, notice that \(\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow \infty}\left(\frac{\frac{1}{x^2}}{1+\frac{1}{x^2}}\right)=\lim_{x \rightarrow \infty}\left(\frac{\frac{1}{x} \frac{1}{x}}{1+\frac{1}{x} \frac{1}{x}}\right)=\left(\frac{\lim_{x \rightarrow \infty}\frac{1}{x} \lim_{x \rightarrow \infty}\frac{1}{x}}{1+\lim_{x \rightarrow \infty}\frac{1}{x} \lim_{x \rightarrow \infty}\frac{1}{x}}\right)=\frac{0}{1}=0\). The last proof is the shortest proof and perhaps the better of the previous two for its brevity.

Because \(f\) is continuous on \([0,\infty)\) and \(\lim_{x \rightarrow \infty}f(x)=0\), \(f\) must be uniformly continuous on \([0,\infty)\). Now since \(f\) is an even function, it is also uniformly continuous on \((-\infty,0]\). Therefore \(f\) is uniformly continuous on \(\mathbb{R}\).