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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.

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Please show how you bounded $\delta_y$ below –  oks Feb 8 '13 at 16:37
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@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. –  mezhang 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. –  mezhang Feb 11 '13 at 19:06
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2 Answers

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}

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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. –  mezhang Feb 11 '13 at 21:16
    
@mezhang have edited to justify $\alpha$ continuous. –  oks Feb 12 '13 at 1:35
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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 $[a,b]$.

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wow great proof! –  mezhang Feb 13 '13 at 20:42
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