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$$f(x) = \begin{cases} x\sin(1/x), & \text{if $x$ $\ne$ $0$} \\ 0, & \text{if $x$ = $0$} \\ \end{cases}$$

Is $f$ continuous on $(-1/\pi$, 1/$\pi$)? Is $f$ differentiable on $(-1/\pi$, 1/$\pi$)?

I know how to prove continuity on a single point, but I'm not sure how to prove continuity for a whole interval. Also, I know there is a theorem that states that if a function is differentiable at a point, then it's continuous.

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  • $\begingroup$ Bolzano's Theorem which is a corollary from the intermediate values theorem. Set $h(x)=f(x)-x$ $\endgroup$ – Haha Oct 15 '14 at 19:13
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    $\begingroup$ An example: $X=[1,2]\cup[-2,-1]$ with $f(x):= -x$? $\endgroup$ – angryavian Oct 15 '14 at 19:14
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If we take as an example $X=[0,1]\cup[2,3]$, $f$ simply changing between these intervals ($f(x)=2+x$ for $x\in [0,1]$, $f(x)=x-2$ for $x\in[2,3]$.

This function is continuous (obviously), and it's range is $X$ (also obviously).

Now, imagine we had a series $x_n$ such that $\lim_{n\to\infty}{(f(x_n)-x_n)}=0$. Since $f$ is continuous, $f-\rm{id}$ is also continuous, and since the domain of $f$ is closed, the range must also be. Thus we know that there would have to be an $x$ with $f(x)-x=0$, which is a contradiction to the first assumption that $f(x)\neq x\quad\forall x\in X$.

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First of all, a good example of $f(x)$ that works for any union $X$ would be $f(x)= x+1$.

Now consider the values of $f(x)-x$ on the two endpoints of each interval. For each interval these values must be non-zero (otherwise $f(x) = x$ on that endpoint). By the intermediate value theorem they must have the same sign on the two endpoints of the interval. And in fact, $f(x)-x$ must be of that same sign throughout the interval, otherwise the IVT tells you that $f(x)-x = 0$ somewhere between the different-sign points.

So for each interval $I_n$ we have a function which is always positive (or always negative) and since this is a closed interval there is a minimum (or for the negative case, a maximum) value of $f(x)-x$ on $I_n$. Note that the fact that the function is continuous on a closed interval is critical here; consider the function $g(x)=x + x^2$ on the open interval $(0,1)$ -- $g(x)-x$ has no minimum on $(0,1)$.

So now taking the set of the absolute values of all those minima and maxima, we have a finite set of numbers, all of which are positive. That set must have a positive minimum (now you know why we needed the word "finite" in the statement of the theorem). And that minimum is the largest $\epsilon$ that works for the theorem.

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    $\begingroup$ Your function will not work for all $X$ (as you claim). Note that the OP requires $f : X \to X$, i.e. the image of $f$ on $X$ has to be a subset of $X$. This will (at least for compact $X$) never be the case for $f(x) = x+1$ (take $x = \max X$). $\endgroup$ – PhoemueX Oct 15 '14 at 21:36
  • $\begingroup$ You are right. I can't find an example that works independent of the union $X$. $\endgroup$ – Mark Fischler Oct 16 '14 at 14:58

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