Uniform continuity is an important concept in real analysis. It's most often introduced after working with the concept of continuity and is a much more stronger statement about a function in the sense that uniform continuity of a real-valued function implies continuity.

What's needed first is the working definition. We say that a function \(f\) is uniformly continuous on \(A \subset \mathbb{R}\) whenever we have that $$(\forall \epsilon >0)(\exists \delta >0)(\forall x_1,x_2 \in A)(|x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|< \epsilon)$$ One very straightforward example of a uniformly continuous function is the real-valued identity function \(f\) with domain \(\mathbb{R}\). For such a function, we let \(\epsilon >0\) be arbitrary, \(\delta=\epsilon\), and \(x_1,x_2 \in \mathbb{R}\). We then have that \(|f(x_1)-f(x_2)|=|x_1-x_2|<\delta=\epsilon\) and therefore the function \(f\) is uniformly continuous.

Our goal will be to prove that \( \frac{1}{x^2+1} \) is uniformly continuous on \(\mathbb{R}\), bringing to the reader's attention various ways of writing proof. The next two posts will help you appreciate discovering more than one way to prove a statement. This first post will develop a theoretical toolbox for showing that \(\frac{1}{x^2+1} \) is uniformly continuous.

Our first tool to introduce is the following statement:

If \(f:[a,\infty) \rightarrow \mathbb{R}\) is continuous and there exists an \(L \in \mathbb{R}\) such that \(f(x) \rightarrow L\) as \(x \rightarrow \infty\), then \(f\) is uniformly continuous on \([a,\infty]\).

Let \( \epsilon > 0\). For this specific \(\epsilon\) we need to show there is a response \(\delta>0\) such that for any two elements \(x,y \in [a,\infty) \) within \(\delta\) apart from each other, \(|f(x)-f(y)|<\epsilon\). To begin, we know that \( \lim_{x \rightarrow \infty}f(x)=L\) yields a real number \(N\) such that for all \(x \geq N\), \( |f(x)-L|<\frac{\epsilon}{3}\). This gives way to three cases that need to be considered. The first case is when the two elements \(x,y\) of the domain are in \([a,N]\), the second when they're in \([N,\infty)\), and the third when one is in \([a,N]\) and the other in \([N,\infty)\) not both equal to \(N\). For each case, it needs to be demonstrated that \(f\) is uniformly continuous.

The first case follows from knowing that continuous functions on a compact set are uniformly continuous. Hence \(f\) is uniformly continuous on \([a,N]\). This yields the result of there being a \(\delta > 0\) for which \(|f(x)-f(y)|<\frac{\epsilon}{3}\) whenever \( |x-y|<\delta\) for any \(x,y \in [a,N]\).

For the second case we have that \(x,y\geq N\) implies $$|f(x)-f(y)|=|f(x)-L+L-f(y)|\leq|f(x)-L|+|f(y)-L|<\epsilon$$ Taking our \(\delta\) from the first case, it follows that for \(x,y\geq N\) and \(|x-y|<\delta\), \(|f(x)-f(y)|<\epsilon\). This establishes uniform continuity of \(f\) on \([N,\infty)\).

Finally we consider the last case whereby we have without loss of generality that \( x<N<y \). Using our same \(\delta\) value from the first case, we then have that for \(|x-y|<\delta\), \(|f(x)-f(y)|=\) $$|f(x)-f(N)+f(N)-L+L-f(y)|\leq$$ $$|f(x)-f(N)|+|f(N)-L|+|L-f(y)|<$$ $$\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$ Having shown all three cases, it is now established that for all \(\epsilon >0\), there is a \(\delta>0\) such that for all \(x,y \in [a,\infty)\), if \(|x-y|<\delta\), then \(|f(x)-f(y)|<\epsilon\).

Therefore, if \(f\) is continuous on \([a,\infty)\) and \(\lim_{x\rightarrow \infty}f(x)\) exists and equal to \(L\), then \(f\) is uniformly continuous on \([a,\infty)\).

Another tool needed has to do with a function being Lipschitz.

We say a real-valued function \(f\) is Lipschitz if for all \(x \neq y\) in the domain of \(f\) we have that \(\left|\frac{f(x)-f(y)}{x-y}\right|\leq M\) for some bound \(M>0\).

An additional tool is stated as follows:

If a real-valued function \(f\) is Lipschitz, then it is uniformly continuous

Let \(\epsilon >0\) and \(f\) be defined from \(A \subset \mathbb{R}\) to \(\mathbb{R}\). From \(f\) being Lipschitz, we have that \(\left|\frac{f(x)-f(y)}{x-y}\right|\leq M\) for all \(x,y \in A\), where \(M>0\). That is the same as saying \(\left|f(x)-f(y) \right| \leq M\left|x-y\right|\). Observe now that \( |f(x)-f(y) |<\epsilon\) for whenever \(|x-y|<\frac{\epsilon}{M}\). Hence \(f\) is uniformly continuous.

The proof for a Lipschitz function being uniformly continuous easily followed from the definition. Proving that an even function uniformly continuous on \( [0,\infty) \) is also uniformly continuous on \( (-\infty,0]\) follows easily too from the definition. Indeed, for \(x,y \in [0,\infty)\), \( |f(-x)-f(-y)|=\left|f(x)-f(y) \right|<\epsilon \).

One last tool to be used in one of the proofs for \(\frac{1}{x^2+1} \) being uniformly continuous uses the Mean Value Theorem. We state it now without proof.

If \(f:[a,b]\rightarrow \mathbb{R}\) is continuous on its domain and differentiable on \((a,b)\), then there exists a point \(c \in (a,b)\) such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

A follow-up post will follow in a few days with the various proofs for \(\frac{1}{x^2+1} \) being uniformly continuous.