uniform continuity

For a uniformly continuous function, when we draw a reactangle around each point with a width slightly less than \[2\delta\] and a height slightly less than \[2\epsilon\], the graph lies completely inside the height of the rectangle.

For functions that aren't uniformly continuous, when we draw a similar rectangle, there will be a segment at some point on the graph where it goes beyond the boundaries of the rectangle.

Intuitively, uniform continuity over an interval can be thought as, "it is guaranteed that when \[x\] changes by a little, \[f(x)\] will also change by a little".

Let \[f:\mathbb{R}\to\mathbb{R}\] be defined on \[S\subseteq \mathbb{R}\]. Then \[f\] is uniformly continuous on \[S\] if for each \[\epsilon>0\] there is a \[\delta>0\] such that \[x,y\in S\] and \[\left| x-y \right|<\delta\] then \[\left| f(x)-f(y) \right|<\epsilon\].

\[f:\mathbb{R}\to\mathbb{R}\] represents a function whose input and output are both real numbers, while \[S\subseteq \mathbb{R}\] means \[S\] is a subset of or equals to set \[\mathbb{R}\]. Combined together, would tell us that \[f\] is a real function in which its \[x\]-values is a subset (or the entirety) of real numbers.

If function \[f\] is a real continuous function on the closed interval \[\left[ a,b \right]\], then \[f\] is uniformly continuous on \[\left[ a,b \right]\].

Assume that \[f\] is not uniformly continuous on \[\left[ a,b \right]\]. Then there is an \[\epsilon>0\] such that for each \[\delta>0\], \[\left| x-y \right|<\delta\] and \[\left| f(x)-f(y) \right|<\epsilon\] fails to hold. Therefore, this implies that for each \[\delta>0\] there exists at least a pair of points \[x,y\in \left[ a,b \right]\] such that \[\left| x-y \right|<\delta\] but \[\left| f(x)-f(y) \right|\ge \epsilon\].

In other words, we're assuming that even if two points (\[x\] and \[y\]) are very close, their \[f\]-values can still be separated by at least \[\epsilon\].

Since the only constraint for \[\delta\] is being positive, we will give \[\delta\] the value of \[\frac{1}{n}\] for each \[n\in\mathbb{N}\] (therefore \[\delta\] will always be positive). Additionally, for each \[n\in\mathbb{N}\], there exists a pair \[x_{n},y_{n}\in \left[ a,b \right]\] so that \[\left| x_{n}-y_{n} \right|<\frac{1}{n}\] but \[\left| f(x_{n})-f(y_{n}) \right|\ge \epsilon\]. Do note that \[x_{n}\] and \[y_{n}\] can be any number within the interval, so for any \[\frac{1}{n}\], you are guaranteed to be able to find something that's smaller than that (due to the infinite density of real numbers).

We can now form an infinite sequence, \[\left( x_{n} \right)\] and \[\left( y_{n} \right)\].

By the Bolzano-Weierstrass theorem, there exists a subsequence, \[\left( x_{n_{k}} \right)\], in the sequence \[\left( x_{n} \right)\] that converges to a single point. Call this point \[x_{0}\]. It follows that \[x_{0}\in \left[ a,b \right]\].

As \[x_{n_{k}}\] approaches \[x_{0}\] (\[\lim_{k\to\infty}x_{n_{k}}=x_{0}\]), the value of \[\delta\], \[\frac{1}{n_{k}}\], approaches \[0\] (\[\lim_{k\to\infty}\frac{1}{n_{k}}=0\]). Clearly, for the condition \[\left| x_{n_{k}}-y_{n_{k}} \right|<\frac{1}{n_{k}}\] to stay true, \[y_{n_{k}}\] must also approach \[x_{0}\].

Since \[f\] is established to be continuous across the interval \[\left[ a,b \right]\], it is guaranteed to be continuous at \[x_{0}\]. This implies that \[f(x_{0})=\lim_{k\to\infty}f \left( x_{n_{k}} \right)=\lim_{k\to\infty}f \left( y_{n_{k}} \right)\], and as a consequence \[\lim_{k\to\infty}\left[ f \left( x_{n_{k}} \right)-f \left( y_{n_{k}} \right) \right]=0\].

However, we've also assumed that \[\left| f(x_{n_{k}})-f(y_{n_{k}}) \right|\ge \epsilon\] for all \[k\], which leads to a contradiction.

On one hand, we have the condition that \[\left| f(x_{n_{k}})-f(y_{n_{k}}) \right|\ge \epsilon>0\] for all \[k\]. On the other, the condition \[\left| f(x_{n_{k}})-f(y_{n_{k}}) \right|\to0\] as \[k\to\infty\]. A sequence of numbers simply cannot stay at least some distance, \[\epsilon\], away from 0 and yet get closer and closer to 0. That is a direct logical fallacy.

Thus, we can conclude that a continuous real function must be uniformly continuous across a bounded interval.

Proofs for uniform continuity usually will have the form:
Choose \[\epsilon>0\]. Let \[\delta=\delta(\epsilon)\] (\[\delta\] be a function of \[\epsilon\]). Choose \[x_{0}\in S\], \[x\in S\]. Assume \[\left| x-x_{0} \right|<\delta\] … therefore \[\left| f(x)-f(x_{0}) \right|<\epsilon\]. The expression for \[\delta\] can only involve \[\epsilon\] and must not involve both \[x\] and \[x_{0}\].

Let \[S=\mathbb{R}\] and \[f(x)=3x+7\]. Prove \[f\] is uniformly continuous on \[S\].

We'll first start with the difference \[\left| f(x)-f(x_{0}) \right|\]: \[\left| \left( 3x+7 \right)-\left( 3x_{0}+7 \right) \right|=3 \left| x-x_{0} \right|\]. Since we want this difference to be smaller than \[\epsilon\], \[3 \left| x-x_{0} \right|<\epsilon\implies \left| x-x_{0} \right|<\frac{\epsilon}{3}\].

Naturally, we would pick \[\delta=\frac{\epsilon}{3}\] to ensure that \[\left| f(x)-f(x_{0}) \right|=3 \left| x-x_{0} \right|<3\delta=3 \left( \frac{\delta}{3} \right)=\epsilon\].

Thus, this chosen \[\delta=\frac{\epsilon}{3}\] will hold for all \[x,x_{0}\in\mathbb{R}\].

The hallmark of uniform continuity is exactly that we can find a \[\delta\] (depending only on \[\epsilon\]) that works no matter which \[x_{0}\]​ we choose. Since we've demonstrated that we can find a \[\delta\] value that works for every \[x_{0}\] and does not depend on neither \[x\] and \[x_{0}\], therefore it is proven that \[f(x)=3x+7\] is uniformly continuous.