08 Continuity and Uniform Continuity
Continuity is often defined in terms of limits, but continuity is actually the easier concept to work with, and it can be defined directly with a definition similar to the definition of limit. There are several equivalent ways to express the continuity of a function at a point.
Definition: Let $a\in \R$, and suppose that $f(x)$ is a function defined on an open interval containing $a$ (including at $a$). Then $f$ is continuous at $a$ if it satisfies any, and hence all, of the following equivalent conditions:
- for any $\eps>0$, there is a $\delta>0$ such that for all $x$, if $|x-a|<\delta$ then $|f(x)-f(a)|<\eps$
- $\ds\lim_{x\to a}f(x)$ exists and $\ds\lim_{x\to a}f(x)=f(a)$
- $\ds\lim_{h\to 0}f(a+h)$ exists, and $\lim_{h\to 0}(f(a+h)-f(a)) = 0$.
It is also convenient to define continuity from the left and continuity from the right at $a$ in the obvious way, using the limits of $f(a)$ from the left and from the right at $a$.
We then define continuity of $f(x)$ on an open interval $(a,b)$ to mean continuity at every point in $(a,b)$, and continuity on a closed interval $[a,b]$ to mean continuity on the open interval $(a,b)$ plus continuity from the left at $a$ and continuity from the right at $b$.
Rules for continuity of the sum, product, and so on follow easily from the corresponding rules for limits. There is also an important new rule for continuity of composition of functions.
Theorem: Let $f$ and $g$ be functions, and $a\in\R$. Suppose that $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $f(a)$. Then $f\circ g$ is continuous at $a$.
Continuity on a bounded, closed interval has several important consequences, including the Intermediate Value Theorem (IVT) and the Extreme Value Theorem (EVT). The IVT seems clearly to depend on the completeness (or "no-holes") property of $\R$, and it can be proved using the least upper bound axiom.
Theorem (Intermediate Value Theorem) Suppose that $f(x)$ is a continuous function on the closed, bounded interval $[a,b]$. If $y$ is some number satisfying $f(a)<y<f(b)$, then there is some $c\in(a,b)$ such that $f(c) = y$. The same holds if $f(a)>y>f(b)$.
Proof: We prove the first assertion; the second is similar. Define $X=\{x\in[a,b]\,|\,f(x)<y\}$. We know that $a\in X$, by the assumption that $f(a)<y$. And $X$ has upper bound $b$ by its definition. So, by the Least Upper Bound Axiom, $X$ has a least upper bound, $\lambda$. Note that $\lambda\in[a,b]$. We show $f(\lambda)=y$ by proving that $f(\lambda)<y$ and $f(\lambda)>y$ are both impossible.
Suppose that $f(\lambda)<y$. So, $y-f(\lambda)>0$. Note that since $f(\lambda)<y<f(b)$, we know $\lambda<b$ and $f$ is continuous from the right at $\lambda$. Since $f$ is continuous from the right at $\lambda$, then (taking $\eps=y-f(\lambda)$ in the definition of continuity) there is a $\delta>0$ such that for all $x$, if $\lambda<x<\lambda+\delta$, then $|f(x)-f(\lambda)|<y-f(\lambda)$. This is equivalent to $-(y-f(\lambda))<f(x)-f(\lambda)<y-f(\lambda)$. In particular, adding $f(\lambda)$ to the second inequality, $f(x)<y$ and therefore $x\in X$. Taking any particular $x\in(\lambda,\lambda+\delta)$, the fact that $x\in X$ contradicts the fact that $\lambda$ is an upper bound for $X$. [The picture here is that if $f(\lambda)<y$, then $f(x)$ is pulled less than $y$ for some $x$'s that are bigger than $\lambda$ and those $x$'s are in $X$, which is impossible since they are bigger than $lub(X)$.]
Now suppose that $f(\lambda)>y$. So, $f(\lambda)-y>0$. Note that since $f(\lambda)>y>f(a)$, we know $\lambda>a$ and $f$ is continuous from the left at $\lambda$. Therefore, there is some $\delta>0$ such that for all $x$, if $\lambda-\delta<x<\lambda$, then $|f(x)-f(\lambda)|<f(\lambda)-y$, which is equivalent to $-(f(\lambda)-y)<f(x)-f(\lambda)<f(\lambda)-y$. In particular, using the first inequality, $f(x)>y$. This means $x\notin X$ for all $x\in(\lambda-\delta,\lambda)$. But that contradicts the fact that $\lambda=lub(X)$ (since it means that there is no element of $X$ greater than $\lambda-\delta$). [The picture here is that if $f(\lambda)>y$, then $f(x)$ is pulled greater than $y$ for all $x$'s just to the left of $y$, which means any of those $x$'s would be a smaller lower bound for $X$.]
The Extreme Value Theorem could be proved directly from the Heine-Borel Theorem, but the textbook chooses to use uniform continuity, which is itself an important new concept. Uniform continuity on an interval is a stronger form of continuity that requires that, for a given $\eps>0$, the same $\delta$ will work for all the points in the interval in the definition of continuity.
Definition: Suppose that the function $f$ is defined on an interval $I$. We say that $f$ is uniformly continuous on $I$ if for every $\eps>0$, there is a $\delta>0$ such that for any $x,y\in I$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\eps$.
The big theorem is that continuity on a closed, bounded interval implies uniform continuity on that interval. The proof that is given in the textbook uses the Heine-Borel Theorem, and it is similar to other proofs that we have seen that use that theorem. The EVT can then be proved using uniform continuity (again, you can refer to the proof in the textbook, where the EVT is called the Max-Min Theorem).
Theorem: Let $f$ be a continuous function on a closed, bounded interval $[a,b]$. Then $f$ is uniformly continuous on $[a,b]$.
Theorem (Extreme Value Theorem): Let $f$ be a continuous function on a closed, bounded interval $[a,b]$. Then there are points $m, M\in [a,b]$ such that for all $x\in [a,b]$, $f(m)\le f(x)\le f(M)$.
The function values $f(m)$ and $f(M)$ are the minimum and maximum values of $f$ on the interval $[a,b]$. The content of the theorem is that, first of all, the function is bounded on $[a,b]$ so that the set of function values $\{f(x)\,|\,x\in[a,b]\}$ has a greatest lower bound and a least upper bound, and second that the function actually achieves those bounds. Note that, in general, when $glb(X)\in X$, then we can refer to the $glb(X)$ as the minimum of $X$, and when $lub(X)\in X$, then we can refer to the $lub(X)$ as the maximum of $X$.