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# real analysis - Construct a function that is nowhere differentiable.

I have been working on this question for a very long time now and seem to have reached a dead end, I will show all my attempted solutions, and any help on the various parts of the question would be much appreciated. Now for Part $(a)$, the graphs I got are the following ; $f_{0}(x)=\sum_{r=0}^{0}2^{-r}K(2^{r}x)$: $f_{1}(x)=\sum_{r=0}^{1}2^{-r}K(2^{r}x)$ $f_{2}(x)=\sum_{r=0}^{2}2^{-r}K(2^{r}x)$ $f_{3}(x)=\sum_{r=0}^{3}2^{-r}K(2^{r}x)$ $f_{10}(x)=\sum_{r=0}^{10}2^{-r}K(2^{r}x)$ Now it is obvious to see that the larger $n$ become the more complicated and horrible looking $f_{n}$ becomes, and to me it looks very discontinuous, even though I know its continuous,
For $(b)$
I feel quite lost as to how to approach this, I tried to decipher the hint , now if $f_{n_{0}}$ is continuous, by definition

For all $\epsilon>0$ $\exists$ $\delta>0$ such that $|x-c|<\delta$ implies $|\sum_{r=0}^{n_{0}}2^{-r}K(2^{r}x)-\sum_{r=0}^{n_{0}}2^{-r}K(2^{r}c)|<\epsilon$
Now I don't know how to go further to incorporate the hint.

1. 2019-11-14

There must be something wrong with the way you generate your graphs. The ones for $f_1$ upto $f_{10}$ are wrong.

For example, already for $f_1$ something goes wrong when the graph shows $f_1(\frac12)= 1$. It should be $\frac12$ because $$f_1(\tfrac12) = K(\tfrac12)+2^{-1}K(2^1\times \tfrac12) = K(\tfrac12) = \tfrac12$$ because $K(1)=0$ so the second term disappears.

Perhaps you've mistyped something such that the by-cases definition of $K(2^rx)$ tests the fractional part of $x$ instead of the fractional part of $2^rx$?

2. 2019-11-14

If you use your first function, f0, and instead of making it look discontinued. Make smaller intervals, make your intervals so small, you will get something like this:

lim
interval --> 0


That way you get the same as in f0, but looking at it from a very far distance (zoomed out). Then your function is still continuous, but cannot be differentiated, I suppose.