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## Proving Uniform Continuity

### By Maurice Ticas

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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 $$\mathbb{R}$$.

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