Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $h : \mathbb{R} \to \mathbb{R}$ be defined by: $$h(t) = \sum_{r(k) < t} \left(\frac{1}{2}\right)^{k},$$ Where $\{r(1) , r(2),\dots\}$ is an enumeration of rational numbers.

Find the points of differentiability of $h$ .

I have found that it is a subset of $\mathbb{R}\setminus\mathbb{Q}$ , but cannot proceed further .

share|improve this question
add comment

1 Answer

up vote 6 down vote accepted

The set of differentiability will depend on the enumeration. Indeed, we can make $h$ differentiable at a given irrational point $\alpha$ by ensuring that $$|r(k)-\alpha|\ge \frac{1}{k} \quad \text{ for all }k$$ This requirement slightly restricts our choice of $r(k)$ but still allows for enumeration of all rationals. To see that $h'(\alpha)=0$, notice that $h(\alpha+1/k)-h(\alpha)\le \sum_{j=k}^\infty 2^{-j}=2^{1-k}$ is much smaller than $1/k$, and similarly for $h(\alpha)-h(\alpha-1/k)$.

On the other hand, we can also make $h$ non-differentiable at a given irrational point $\alpha$ by choosing $|r(k)-\alpha|<4^{-k}$ when $k$ is even, and using the odd indices to enumerate the rest of rationals. With this enumeration $h(\alpha+4^{-k})-h(\alpha)>2^{-k}$, which is much larger than $4^{-k}$. Hence $\limsup_{x\to \alpha}\frac{h(x)-h(\alpha)}{x-\alpha}=\infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.