Continuous Functions on a Closed Interval: Uniform Continuity

Uniform Continuity vs. Continuity

We have discussed some very useful properties of continuous functions in the last few posts. In the current post we will focus on another property called "uniform continuity". To understand what this is all about it first makes sense to reiterate the meaning of usual concept of continuity as we have seen earlier.

A function $ f$ is a said to be continuous at a given point $ x = a$ if it is defined in a neighborhood of $ x = a$ and if for any arbitrarily specified number $ \epsilon > 0$ we can find another number $ \delta > 0$ such that $ |f(x) - f(a)| < \epsilon$ whenever $ |x - a| < \delta$.

In the above definition it should be noted that in general the value of number $ \delta$ will depend upon the value of $ \epsilon$. Moreover it is not stated explicitly but the value of $ \delta$ will also depend upon the point $ x = a$ under consideration. To make thing clear it is better to take a simple example. Let $ f(x) = 1/x$ for $ x > 0$. And we will check it continuity at two points $ x = 10$ and $ x = 1/10$. To ensure continuity of $ f(x)$ at $ x = 10$ we must be able to find $ \delta > 0$ for given $ \epsilon > 0$ such that $ |f(x) - f(10)| < \epsilon$ whenever $ |x - 10| < \delta$.

Now $$ |f(x) - f(10)| = |(1/x) - (1/10)| = |(x - 10)/(10x)| < |(x - 10)| / 90$$ because we can take $ x > 9$ (this means that $ \delta \leq 1$). Now it is clear that if we take $ \delta = 90\epsilon$ and also ensure $ \delta < 1$, the definition of continuity is satisfied.

When we analyze the continuity of $ f(x)$ at $ x = 1/10$. We have to ensure $ |f(x) - f(1/10)| < \epsilon$ by taking $ |x - (1/10)| < \delta$. Since $$ |f(x) - f(1/10)| = |(1/x) - 10| = |(1 - 10x)/x| < 11|10x - 1|$$ as we can take $ x > 1/11$ (here it requires $ \delta \leq 1/10 - 1/11 = 1/110$). Now $ 11|10x - 1| < \epsilon$ if $ 110|x - (1/10)| < \epsilon$, so that we can take $ \delta = \epsilon / 110$. Hence finally $ \delta$ can be taken as smaller of the two numbers $ 1/110$ and $ \epsilon/110$.

The above example shows that the for the same value of $ \epsilon$ we may have to choose different values of $ \delta$ depending upon the point of continuity under consideration. This is only to be expected because the function may change its values sharply in neighborhood of one point and around another point the changes in values of the function might not be that great.

However there can be functions for which the same value of $ \delta$ may suffice for all points under consideration. For example $ f(x) = x$ clearly has this property (taking $ \delta = \epsilon$ suffices to ensure continuity at every point). Such functions we call as "uniformly continuous".

A formal definition is as follows:
Let $ f$ be a function defined on an interval $ I$. The function $ f$ is said to be uniformly continuous on interval $ I$ if for any specified number $ \epsilon > 0$ it is possible to find a corresponding number $ \delta > 0$ such that for any points $ x, y \in I$ we have $ |f(x) - f(y)| < \epsilon$ whenever $ |x - y| < \delta$.

From the definition we can observe that uniform continuity is a global property (related to the behavior of a function in an interval) in contrast with the usual continuity which is a local property (related to the behavior of a function around one single point under consideration). Moreover it is quite obvious from the above definition that if $ f$ is uniformly continuous on an interval $ I$, then it is also continuous on the the interval $ I$.

But the reverse may not hold. The function $ f(x) = 1/x$ is continuous on interval $ (0, 1)$ but not uniformly continuous on $ (0, 1)$. It turns out that if we restrict ourselves to closed intervals both the concepts of continuity turn out to be equivalent.

Functions continuous on a closed interval are uniformly continuous on the same interval.

Here we have to establish a global property from a local one and hence it is apparent that Heine Borel Principle should help us out. So let $ f$ be continuous on a closed interval $ [a, b]$ and define $ f(x) = f(a)$ for $ x < a$ and $ f(x) = f(b)$ for $ x > b$ so that $ f(x)$ is continuous everywhere. For any specified number $ \epsilon > 0$ and any given point $ x \in [a, b]$ we can find a neighborhood $ I_{x} = (x - \delta_{x}, x + \delta_{x})$ such that $ |f(y) - f(x)| < \epsilon / 2$ whenever $ y \in I_{x}$. The collection of all these neighborhoods $ I_{x}$ forms a cover of $ [a, b]$ and hence we can choose a finite number of these neighborhoods, say $ I_{x_{1}}, I_{x_{2}}, \cdots, I_{x_{n}}$ to cover the interval $ [a, b]$.

The end points of these chosen neighborhoods which fall in the interval $ [a, b]$ divide the interval $ [a, b]$ into a finite number of sub-intervals, say $ I_{1}, I_{2}, \cdots, I_{m}$. Let $ \delta$ be any positive number smaller than the length of smallest of these sub-intervals ($ I_{1}, I_{2}, \cdots, I_{m}$). Now if we choose any two points $ x, y \in [a, b]$ with the condition $ |x - y| < \delta$ then it is clear that both these points must lie in one of the neighborhoods $ I_{x_{j}}$. Hence
\begin{align}|f(x) - f(y)| &= |f(x) - f(x_{j}) + f(x_{j}) - f(y)|\notag\\ &\leq |f(x) - f(x_{j})| + |f(y) - f(x_{j})|\notag\\ &< \epsilon/2 + \epsilon/2\notag\\ &= \epsilon\notag\end{align} It follows that the function $ f$ is uniformly continuous on $ [a, b]$.

Without this (or an equivalent) property of continuous functions it is almost impossible to establish the integrability of continuous functions (because integration again is considered on intervals and therefore requires analysis of global properties of the function).

Why Closed Intervals?

A few readers must be wondering why all these beautiful properties of continuous functions are established only for closed intervals. Why don't they hold for open intervals? To understand this we need to look at the methods of proof used in establishing these properties. In general if we are using the Supremum principle (or the equivalent Dedekind's Theorem) we need to have the behavior of function known at the end points of the interval too. Since the open interval does not contain the end-points it just isn't possible to start the proof along these lines.

A second approach was based on the Nested Interval Principle. Clearly this also fails for the case of open intervals. As an example let us consider the sequence of intervals $ I_{n} = (0, 1/n)$. They form a nested sequence, but there is no point which is common to all the intervals. For if there were a point common to all these intervals, it must be a positive number say $ x$. Now it needs to be in all the intervals so we must have $ x < 1/n$ for all values of $ n$ which is false as $ x$ is a fixed positive number. The reason the Nested Interval Principle fails for open intervals is that the sequence of their end points may reach (in the limit - to be explained a bit later) outside the intervals being considered (here in the example $ 1/n$ reaches $ 0$ in the limit). If the intervals included these limiting points also the Nested Interval Principle would remain valid.

Again lets discuss something about the Heine Borel Principle. Clearly if we have an open interval then we don't have end points which are covered. So there may be cases where the finite number of intervals chosen to cover $ (a, b)$ will not be able to include points which are very close to the end point $ a$ or $ b$. Again it is to be noted that the reason the principle fails is that we can find a sequence of points $ x_{n}$ in $ (a, b)$ such that their limit itself becomes $ a$ or $ b$. In other words the open intervals do not contain the limit points.

The deficiency of open intervals was being discussed crudely in terms of some points being termed as limit points. It's time now to make a formal definition.
Let $ A$ be a non-empty set of real numbers. If there exists a real number $ c$ (which may or may not belong to the given set $ A$) with the property that for any given number $ \epsilon > 0$ there exists at least one member $ x \in A$ such that $ 0 < |x - c| < \epsilon$, then $ c$ is said to be an accumulation point or a limit point of the set $ A$.

It should be observed that if a set $ A$ has a limit point $ c$ then we must be able to find members of $ A$ which are as close to (but not equal to) $ c$ as we please. Again this means (by repeating the argument) that we should be able to find an infinity of members of $ A$ which are as close to $ c$ as we please. Therefore if $ A$ has a limit point then $ A$ must be an infinite set. In other words finite sets do not have limit points. At the same time it does not mean that any infinite set will necessarily have a limit point. For example the set of integers has no limit point.

Also it must be noted that limit point of a set may or may not belong to the set itself. However it may happen that a set contains all its limit points. Such a set is called a closed set. A closed interval is actually a closed set by this definition (hence the name closed). An open interval misses its two end points $ a, b$ which are limit points of the interval.

Now we can discuss an important theorem which guarantees the existence of limit points under certain conditions.

Bolzano Weierstrass Theorem: An infinite set which is bounded must have a limit point.

Since the infinite set $ A$ is bounded it can be assumed to lie entirely in an interval $ [a, b]$. Clearly since A is infinite, at least one the intervals $ [a, (a + b)/2]$ and $ [(a + b)/2, b]$ must contain an infinite number of members of $ A$. Choose that interval and call it $ [a_{1}, b_{1}]$. Repeating the procedure again and again we get a sequence of intervals $ [a_{n}, b_{n}]$ such that there lie an infinity of members of $ A$ in each of the intervals $ [a_{n}, b_{n}]$. By Nested Interval Principle there is a unique number $ c$ lying in all these intervals (uniqueness comes because of $ b_{n} - a_{n} = (b - a)/2^{n} < (b - a)/n$). This $ c$ must be a limit point of $ A$ because any neighborhood $ (c - h, c + h)$ contains an interval $ [a_{n}, b_{n}]$ and hence an infinite number of points from $ A$.

We can apply Dedekind's Theorem also. Let us divide all real numbers in two sets $ L, U$ in the following way. A number $ x$ lies in $ L$ if it exceeded an infinite number of members of $ A$ and $ x$ lies in $ U$ if it exceeded only by a finite number of members of $ A$. Clearly the above division forms a section of real numbers and therefore there is a number $ c$ such that for any $ h > 0$, $ c - h \in L$ and $ c + h \in U$. Therefore an infinite number of members of $ A$ are greater than $ c - h$ and only a finite number of these are greater than $ c + h$. Therefore $ (c - h, c + h)$ contains an infinite number of members of $ A$ for any $ h > 0$. This proves that $ c$ is a limit point of $ A$.

Now to the original question of importance of closed intervals. If we observe carefully the basic principles of real variable theory like Dedekind's Theorem, Supremum Principle and Nested Interval Principle are in effect trying to find a limit point of the interval under consideration in various different ways. The fact that the interval is closed guarantees that this limit point will also lie in the same interval. The story for Heine Borel Principle is a bit different. In case we have a set $A$ which is not closed (take for example a finite open interval) and we try to cover it with a finite number of sub-intervals we may miss the limit points of $A$ which are lying outside $A$ and thereby miss the points which are lying near to this limit point but lying in $A$. This can happen when the intervals covering these points are small enough that they don't cover the nearby limit point.

For example consider interval $ I = (0, 1)$ Clearly for each $ x \in I$ we have a an interval $ I_{x} = (x/2, 3x/2)$ which includes the point $ x$. The collection of all these intervals forms a cover for $ I$. If we take any finite number of intervals from this collection say $ I_{x_{1}}, I_{x_{2}}, \ldots,I_{x_{n}}$ and let's keep $ x_{1}$ to be the least of all numbers $ x_{1}, x_{2}, \ldots, x_{n}$. Then clearly these finite number of intervals can reach points in the interval $ (x_{1}/2, 1)$ but no points which lie to the left of $ x_{1}/2$. Thus the points of the interval $ (0, x_{1}/2]$ are not covered by this finite collection of intervals. Therefore it is not possible to cover $ (0, 1)$ by any finite number of intervals from the given collection of intervals $ I_{x}$. So the Heine Borel Principle also applies to closed sets only. The boundedness of the set is also essential for two reasons. First the boundedness ensures existence of limit points and second it is necessary because a finite number of intervals can't be guaranteed to cover an unbounded set.

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