Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f: [a,b]\to\mathbb{R}$ be continuous, prove that it is uniform continuous.

I know using compactness it is almost one liner, but I want to prove it without using compactness. However, I can use the theorem that every continuous function achieves max and min on a closed bounded interval.

I propose proving that some choices of $\delta$ can be continuous on $[a,b]$, for example but not restricted to:

For an arbitrary $\epsilon>0$, for each $x\in[a,b]$ set $\Delta_x=\{0<\delta<b-a \;|\;|x-y|<\delta\Longrightarrow |f(x)-f(y)| <\epsilon\}$, denote $\delta_x = \sup \Delta_x $.

Basically $\delta_x$ is the radius of largest neighborhood of $x$ that will be mapped into a subset of neighborhood radius epsilon of $f(x)$. I'm trying to show that $\delta_x$ is continuous on $[a,b]$ with fixed $\epsilon$. My progress is that I can show $\delta_y$ is bounded below if $y$ is close enough to $x$, but failed to find its upper bound that is related to its distance with $x$.

Maybe either you could help me with this $\delta_x$ proof, or another cleaner proof without compactness (but allowed max and min). Thanks so much.

share|cite|improve this question
Please show how you bounded $\delta_y$ below – oks Feb 8 '13 at 16:37
@oks idea is that if y is very close enough to x, f(y) is in epsilon/2 neighborhood of f(x). There is a neighborhood of y, inside a neighborhood of x, where points in this neighborhood will be within epsilon/2 neighborhood of f(x), therefore be in epsilon neighborhood of f(y). That radius of the neighborhood is a lower bound on $\delta_y$. Writing out will use a lot symbols and possibly confusing. – mez Feb 8 '13 at 16:43
Is answer below valid? – oks Feb 11 '13 at 15:42
@oks Looking at it right now, will respond today. – mez Feb 11 '13 at 19:06
up vote 3 down vote accepted

Since $f$ is continuous, one of the end-points of the $\delta_x$ neighbourhood is a value $z$ such that $f(z) = f(x) \pm \epsilon$. Otherwise you could extend the neighbourhood and still be within $\epsilon$ of $f(x)$.

Similarly to your argument in the comment, for $m \ge 1$, the radius of largest neighborhood of $x$ that will be mapped into a subset of neighborhood radius $m \epsilon$ is $$\delta_x (1 + \alpha(m))$$ for some finite $\alpha(m) > 0$, where $\alpha(m) \rightarrow 0$ as $m \rightarrow 1$. So either $$f(x + \delta_x(1 + \alpha)) = f(x) \pm m \epsilon$$ or $$f(x - \delta_x(1 + \alpha)) = f(x) \pm m \epsilon.$$

Pick $y$ sufficiently close to $x$ that $|f(y)-f(x)| < (m-1)\epsilon$ and $0 < |y - x| < \alpha\ \delta_x$. Then the interval $I = [y - \delta_x(1 + 2\alpha), y + \delta_x(1 + 2\alpha)]$ contains both $x + \delta_x(1 + \alpha)$ and $x - \delta_x(1 - \alpha)$, and so $f(I)$ contains a point which is $m \epsilon$ away from $f(x)$, and so is more than $\epsilon$ away from $f(y)$. So $$\delta_y < \delta_x(1 + 2\alpha).$$

As $m \rightarrow 1, \alpha \rightarrow 0$ and $y \rightarrow x$, so $\delta_y \rightarrow \delta_x$.

Edit: to show that $\lim_{m \rightarrow 1} \alpha(m) = 0$ as raised in comment.

At least one of the end-points of the $\delta_x$ neighbourhood is a point $z$ such that $f(z) = f(x) \pm \epsilon$ and $|f - f(x)|$ strictly increases for some non-zero interval outside the neighbourhood. For convenience, suppose the end-point is the right end-point $x + \delta_x$. Then for some $\delta > 0$, by continuity of $f$, \begin{align}\forall w \in (x + \delta_x, x + \delta_x + \delta], \\ |f(w) - f(x)| > \epsilon \\ \mbox{and }w_1 > w_2 \Rightarrow |f(w_1)-f(x)| > |f(w_2) - f(x)|.\end{align}

For simplicity, suppose that only the right end-point of the $\delta_x$ interval bites, and let $\delta$ be small enough that $\forall w \in [x - \delta_x - \delta, x - \delta_x], |f(w) - f(x)| \le \epsilon$. Then for $w \in [x + \delta_x, x + \delta_x + \delta]$, $w$ uniquely defines the border of the $\delta_x(1 + \alpha(m))$ neighbourhood of $x$ where \begin{align} m = \frac{f(w) - f(x)}{\epsilon} \\ \mbox{and }w - x = \delta_x(1 + \alpha(m)) \\ \Rightarrow \alpha(m) = \frac{w - x}{\delta_x} -1. \end{align}

Since $m$ and $\alpha(m)$ are continuous functions of $w$, $\alpha(m)$ is a continuous function of $m$ i.e. given $\eta > 0$ you can choose small enough $w \in [x + \delta_x, x + \delta_x + \delta]$ such that for all $v \in [x + \delta_x, w), $

\begin{align} 1 < m(v) < \frac{f(w) - f(x)}{\epsilon} \\ \mbox{and } 0 < \alpha(m(v)) < \frac{w - x}{\delta_x} -1 < \eta\\ \Rightarrow \lim_{ m\rightarrow 1} \alpha(m) = 0.\end{align}

share|cite|improve this answer
Everything is clear, but how do you justify $\alpha$ is continuous on $m$? $m = 1, \alpha = 0$ is clear, but that doesn't mean $\displaystyle\lim_{m\to 1} \alpha(m)$ exists. – mez Feb 11 '13 at 21:16
@mezhang have edited to justify $\alpha$ continuous. – oks Feb 12 '13 at 1:35

Let an $\epsilon>0$ be given and put $$\rho(x):=\sup\bigl\{\delta\in\ ]0,1]\ \bigm|\ y, \>y'\in U_\delta(x)\ \Rightarrow\ |f(y')-f(y)|<\epsilon\bigr\}\ .$$ By continuity of $f$ the function $x\to\rho(x)$ is strictly positive and $\leq1$ on $[a,b]$.

Lemma. The function $\rho$ is $1$-Lipschitz continuous, i.e., $$|\rho(x')-\rho(x)|\leq |x'-x|\qquad \bigl(x,\ x'\in[a,b]\bigr)\ .$$

Proof. Assume the claim is wrong. Then there are two points $x_1$, $\>x_2\in[a,b]$ with $$\rho(x_2)-\rho(x_1)>|x_2-x_1|\ .$$ It follows that there is a $\delta$ with $\rho(x_1)<\delta<\rho(x_2)-|x_2-x_1|$. By definition of $\rho(x_1)$ we can find two points $y$, $\> y'\in U_\delta(x_1)$ such that $|f(y')-f(y)|\geq\epsilon$. Now $$|y-x_2|\leq |y -x_1|+|x_1-x_2|<\delta +|x_2-x_1| =:\delta'<\rho(x_2)\ .$$ Similarly $|y'-x_2|<\delta'$. It follows that $y$, $\>y'$ would contradict the definition of $\rho(x_2)$.$\qquad\quad\square$

The function $\rho$ therefore takes a positive minimum value $\rho_*$ on $[a,b]$. The number $\delta_*:={\rho_*\over2}>0$ is a universal $\delta$ for $f$ and the given $\epsilon$ on $[a,b]$.

share|cite|improve this answer
wow great proof! – mez Feb 13 '13 at 20:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.