Hello!

I want to know if the following is correct:

I want to prove that the function

$\displaystyle f_{i} (x) = \int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}$ is differentiable on all open intervals $\displaystyle (2^{-i}k, 2^{-i}(k+1))$ with $\displaystyle k \in \mathds{N} \cup \{0 \}$

Here is what I did:

I know that the function is linear on the given open intervals and each linear function is differentiable.

To show that the function is not differentiable on $\displaystyle 2^{i}k$ I calculated using I'Hospital in the differential quotient :

$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{f_{i} (x) - f_{i} (2^{-i} k^{+})}{x-2^{-i} k^{+}} = \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{\int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}- \int_{0}^{2^{-i} k^{+}} (-1)^{\lfloor 2^{i} t\rfloor}}{x-2^{-i} k^{+}}$

I'Hospital gives:

$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{(-1)^{\lfloor 2^{i} x\rfloor}}{1} = \lim\limits_{x \rightarrow 2^{-i} k^{+}} (-1)^{\lfloor 2^{i} x\rfloor} = (-1)^{\lfloor k^{+} \rfloor} = (-1)^k \quad (1) $

On the other hand

$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{-}} \frac{f_{i} (x) - f_{i} (2^{-i} k^{+})}{x-2^{-i} k^{+}} =(-1)^{\lfloor k^{-} \rfloor} = (-1)^{k-1} \quad (2)$

Now when k is even then (1) is 1 and (2) -1 and

if k is odd, then (1) is -1 and (2) is even.

Given that fact, it means that the function $\displaystyle f_i (x)$ is not differentiable at $\displaystyle 2^{i}k$

I would do the same calculation with the 2nd endpoint of the given interval.

My question is: Is it correct what I am doing here?

Thanks.

Regards