Showing the Cantor function is not Lipschitz.

Question

This is one I am having a lot of difficulty with. I'm not sure how to show that the Cantor function (or 'Devil's Staircase) is not Lipschitz.

Maybe you meant isn't Lipschitz? – k.stm Oct 02 '12 at 17:44 — k.stm, Oct 02 '12 at 17:44
You are correct. I need to fix that. – emka Oct 02 '12 at 17:48 — emka, Oct 02 '12 at 17:48

score 8 · Accepted Answer · answered Oct 02 '12 at 17:50

8

Hint: For every nonnegative integer $n$, find some points $x_n$ and $y_n$ such that $|x_n-y_n|=1/3^n$ and $|f(x_n)-f(y_n)|=1/2^n$. Conclude.

answered Oct 02 '12 at 17:50

Did

I'm not sure I follow.
I guess I may be misunderstanding what Lipschitz means. In my mind I like to think of a function having the Lipschitz property as saying that the secants are bounded by a positive M.
– emka Oct 02 '12 at 19:26
Precisely. The hint indicates that the slope of the secant between $x_n$ and $y_n$ is pretty large, when $n$ is large... – Did Oct 02 '12 at 19:29
1

So would this be where you are going:
$$\frac{|f(x_n)-f(y_n)|}{|x_n-x_y|} \leq \frac{3^n}{2^n}$$.

I guess I'm still not sure what there is that $\leq$. In my mind, isn't is exactly equal to $\frac{3^n}{2^n}$.
– emka Oct 02 '12 at 20:26
Yes, equal. $ $ – Did Oct 03 '12 at 04:58
I noticed that my previous comment made marginal sense. What I meant to say is: why is it $\leq$ and not $=$? My gut feeling is to claim this $\leq$, but I don't have a defense for it. – emka Oct 03 '12 at 06:52
If you found $x_n$ and $y_n$, you can see that this is $=$. If you did not, I do not understand this conversation. (Note that $\leqslant$ may hold for functions as regular as one wants, for example constant.) – Did Oct 03 '12 at 09:15

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