# differentiable point set

• Dec 23rd 2009, 02:08 AM
Shanks
differentiable point set
Find a suffice and neccessary condition that a set E in real numbers such that there is a real function whose differentiable point set is exactly E.
• Dec 23rd 2009, 09:26 AM
Jose27
Quote:

Originally Posted by Shanks
Find a suffice and neccessary condition that a set E in real numbers such that there is a real function whose differentiable point set is exactly E.

Since we're talking about differentiability we have that $\displaystyle E$ is open (or are you not allowed to assume this?). If $\displaystyle E$ is open it is the union of disjoint open intervals so in each interval $\displaystyle (a,b)$ you define $\displaystyle f$ as follows: If $\displaystyle z\in (a,b)$ then $\displaystyle f(z)= \sum_{n=1}^{\infty } \left( \frac{z-z_0}{c} \right) ^n$ where $\displaystyle z_0= a+\frac{b-a}{2}$ and $\displaystyle c$ is such that $\displaystyle \vert \frac{z-z_0}{c} \vert < 1$. Then $\displaystyle f$ is defined in every interval and cannot be extended beyond them.

If you're not allowed to assume E open I don't really know how you could approach this...

On a related note, this reminded me of this Domain of holomorphy - Wikipedia, the free encyclopedia in the case of one variable