4
$\begingroup$

My thoughts was to take $f(x) =\cos(\frac 1x) $ for all $ x \in [0,1)$ as I know this function is continous from $[0,1)$ and is definitely not uniformly continuous as it oscilates non-uniformly. My trouble is with the proof.

To prove continuity would I:

Fix $x_0 \in [0,1), \epsilon>0.$ We will show that there exists $\delta>0$ such that if $|x-x_0|<\delta$ then $|\cos(\frac 1x) -\cos(\frac 1{x_0})|<\epsilon$

Now I am stuck as to how I could simplify $|\cos(\frac 1x) -\cos(\frac 1{x_0})|$ or what $\delta$ to choose. Any help would be appreciated.

$\endgroup$
  • 2
    $\begingroup$ You function is actually not defined at $x=0$. $\endgroup$ – Arctic Char Apr 6 at 5:14
  • $\begingroup$ What would you suggest as a function that I could use? $\endgroup$ – Plus Twenty Apr 6 at 5:16
  • $\begingroup$ Why would you consider $\cos (1/x)$ in the first place? $\endgroup$ – Arctic Char Apr 6 at 5:17
  • $\begingroup$ It was the first function I thought of that was continuous but not uniformly continuous. $\endgroup$ – Plus Twenty Apr 6 at 5:18
  • 1
    $\begingroup$ @TheoBendit $\frac1{x-1}$ is unbounded. $\endgroup$ – José Carlos Santos Apr 6 at 5:48
6
$\begingroup$

Here's some intuition:

The Heine-Cantor theorem tells us that any function between two metric spaces that is continuous on a compact set is also uniformly continuous on that set (see here for discussion). Next, if $f:X \rightarrow Y$ is a uniformly continuous function, it is easy to show that the restriction of $f$ to any subset of $X$ is itself uniformly continuous*. Therefore, because $[0,1]$ is compact, the functions $[0,1) \to \mathbb{R}$ that are continuous but not uniformly continuous are those functions that cannot be extended to $[0,1]$ in a continuous fashion.

For example, consider the function $f:[0,1) \to \mathbb{R}$ defined such that $f(x) = x$. We can extend $f$ to $[0,1]$ by defining $f(1) = 1$, and this extension is a continuous function over a compact set (hence it is uniformly continuous). So the restriction of this extension to $[0,1)$—i.e. the original function—is necessarily also uniformly continuous per (*) above.

How can we find a continuous function on $[0,1)$ that cannot be continuously extended to $[0,1]$? There are two ways:

$\qquad \bullet \quad$ Construct $f$ so that $\displaystyle \lim_{x \rightarrow 1} f(x) = \pm \infty$

$\qquad \bullet \quad$ Construct $f$ so that $\displaystyle \lim_{x \to 1} f(x)$ does not exist

Note that if $\displaystyle \lim_{x \to 1} f(x)$ exists, taking $f(1)$ to be that limit yields a continuous extension. Indeed, $\displaystyle \lim_{x \to c} f(x) = f(c)$ is literally one of the definitions for continuity at the point $x=c$.

The first bullet is ruled out by the stipulation that $f$ be bounded, so moving on to the second bullet, we need to make sure $\displaystyle \lim_{x \to 1} f(x)$ does not exist. One way of doing this (the only way I believe) is to have $f$ oscillate infinitely rapidly with non-vanishing amplitude as $x \to 1$. It looks like this is what you were trying to exploit, and what José Carlos Santos did (+1) in his answer (see graph below): $\displaystyle f(x) = \cos \left(\frac{1}{1-x} \right)$.

Generalizing his epsilon-delta argument, you can see that this condition is not only necessary, but also sufficient.

$\qquad \qquad \qquad \qquad$ enter image description here

$\endgroup$
5
$\begingroup$

Take $f(x)=\cos\left(\frac1{1-x}\right)$. If it was uniformly continuous, then, for each $\varepsilon>0$, there would be some $\delta>0$ such that $\lvert x-y\rvert<\delta\implies\bigl\lvert f(x)-f(y)\bigr\rvert<\varepsilon$. But this is not true. Take $\varepsilon=1$. Since there are values of $x$ arbitrarily close to $1$ such that $f(x)=1$ and there are values of $x$ arbitrarily close to $1$ such that $f(x)=-1$, then, no matter how small $\delta$ is, you will always be able to find examples of numbers $x,y\in[0,1)$ such that $\lvert x-y\rvert<\delta$ and that $\bigl\lvert f(x)-f(y)\bigr\rvert=2>\varepsilon$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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