If $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$.

Question

So I want to prove that continuity on $[a,b]$ implies uniform continuity with only using the least upper bound property of the reals. I know the basic idea of this, but am getting confused with choosing the right $\delta$. Here's where I am so far:

Proof. Let $\epsilon >0$ and define $$A(\delta) = \{u \in [a,b] ~| \text{ if } x,t \in [a,u] \text{ and } |x-t| < \delta, \text{ then } |f(x)-f(t)| < \epsilon \},$$ and $$A = \bigcup_{\delta >0} A(\delta).$$ Since $a \in A$ and $b$ is an upper bound for $A$, $\alpha = \sup(A)$ exists. Now I need to show two things: first that $\alpha = b$, and then that $\alpha \in A$. To show $\alpha = b$ assume that $\alpha <b$. Then by continuity there exists some $\delta(\alpha) > 0$ such that if $|x-\alpha|<\delta(\alpha)$, then $|f(x)-f(\alpha)|< \epsilon$. Now since $\alpha = \sup(A)$, there exists some $x_0 \in A$ such that $\alpha - \delta(\alpha) < x_0 \le \alpha$. Then there exists some $\delta(x_0)>0$ such that $x_0 \in A(\delta(x_0))$. Now let $\delta_{\text{min}} = \min\{\delta(x_0), \delta(\alpha)\}$...

So here is where I am stuck. For starters I'm not sure if this $\delta$ will work. Also, I am imagining that I will need to use the triangle inequality to show that $\alpha \in A(\delta^*)$ where $\delta^*$ is whichever $\delta$ that will do the trick, but I'm not sure what to use the triangle inequality on. Basically I've confused myself. Help?

Answer 1

I came across this problem when I reading this undergrad analysis book, and I think this problem is a little bit challenging unless I was overdoing it. Here is my proof:

First we have $A=\bigcup_{\delta>0} A(\delta)$ is bounded above by $b$ by the construction of $A(\delta)$, and it's nonempty because it contains $a$. So there exists $c=l.u.b.(A)$, of course $c\leq b$ since $b$ is an upper bound.

Some fact: (1) if $u \in A(\delta)\subset A$, then for any $v\in[a,u)$, $v$ is also in the same $A(\delta)$, therefore in $A$. This is just saying that if there is any $u\in A$, then $[a,u]\subset A$. Proof: play with the definition.(omitted)

(2)For fixed $\epsilon$, if $s\in A$ where A is induced by $\epsilon$, then $s\in A$ where A is induced by $\frac{\epsilon}{2}$. Proof: rescale $\epsilon$.

Since function $f$ is continuous at $c$, for any $\epsilon >0$,there exists $\delta_0>0$ such that for any $y$ that satisfies $|y-c|\leq\delta_0$ we have $|f(c)-f(y)|\leq \epsilon$. Now we pick this very $\delta_0$, since $c$ is the l.u.b.(A) so there exists $s\in A$ such that $c-\delta_0\leq s\leq c$, since $s\in A=\bigcup_{\delta>0}A(\delta)$, so there exists $\delta_1$ such that $s\in A(\delta_1)$. Now we want to show that $c$ is also in $A$:

Fact (1) tells us $[a,s]\subset A(\delta_1)$, let $\delta=min\{\delta_0, \delta_1\}$. Then for any $x,t \in [a, c]$ with $|x-t|<\delta$,

case1: If $x,t$ are both in $[a,s]$, then $|f(x)-f(t)|<\epsilon$ because $f$ is uniformly continuous on $[a,s]$.

case2: If $x,t$ are both in $[s, c]$, then $|f(x)-f(t)|\leq |f(x)-f(c)|+|f(c)-f(t)|<\epsilon + \epsilon=2\epsilon$.

case3: If one is in $[a,s]$ and the other is in $[s,c]$. WLOG, let $x\in [a,s]$ and $t\in [s,c]$, then $|f(x)-f(t)|\leq |f(x)-f(s)|+|f(s)-f(c)|+|f(c)-f(t)|\leq \epsilon + \epsilon +\epsilon =3\epsilon$.

Anyway, we can use fact (2) to rescale and get a uniform $\epsilon$. So we've shown $c$ is also in $A$. Now if we assume $c<b$, then we can repeat the above argument to show $c+\delta_0$ is also is also in $A$ (use $c$ as the intermediate point), contradicting the fact the $c$ is the l.u.b.(A). So the only possible way is that $c=b$ and $c+\delta_0$ is out of the bound of domain $[a,b]$. Therefore $c=b\in A$, done.

P.S.: I've carefully checked my reasoning and it should be right up to some detail work, e.g. rescaling issue. And I'm pretty sure there is a better way to put all these together nicely, but I'm out of time so...