Usbrth

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 $|cos(frac 1x) -cos(frac 1x_0)|<epsilon$

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

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 $[0,1) to mathbbR$ 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 $f:[0,1) to mathbbR$ 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): $displaystyle f(x) = cos left(frac11-x right)$.

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

$qquad qquad qquad qquad$

Take $f(x)=cosleft(frac11-xright)$. If it was uniformly continuous, then, for each $varepsilon>0$, there would be some $delta>0$ such that $lvert x-yrvert<deltaimpliesbigllvert f(x)-f(y)bigrrvert<varepsilon$. 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 $x,yin[0,1)$ such that $lvert x-yrvert<delta$ and that $bigllvert f(x)-f(y)bigrrvert=2>varepsilon$