# Analyticity of a real function on $[0,\infty)$

I'm struggling to understand the difference of the analyticity of a real and a complex functions. Consider the following real valued function which is a minimal example of a somewhat more involved problem. I'm only interested in the behavior on the real axis in the interval $[0,\infty)$ and for $b>1$

$$f(x)=\exp\left( -x^b\right)$$

Since the composition of two analytic functions is again holomorphic, I guess we could reduce this further, leave out the exponential and concentrate on the power function. For this, a related problem was already given in this question although there, the goal was to investigate in the complex plane.

Question: What can be said about the analyticity of $f$ especially at $x=0$ when the real valued $b>1$?

### Some observations and side questions

I chose $b>1$ to ensure that the derivative exists at $x=0$. If we derive we get

$$\partial_x \,f(x)=-bx^{b-1}\cdot \exp\left(-x^b\right)$$

and choosing $b\leq1$ would lead in the limit $x\to0$ to either $0^0$ or infinity. This doesn't seem to solve anything because even when $b>1$ I cannot calculated the limit $x\to-0$ because the expression is complex (and not real valued) for $x<0$. This leads to some side questions which I really like to have an answer for:

1. Can one say the first derivative of $f$ exist at $r=0$, although I cannot take the limit from the right side?

2. If I only need the function for $x\in[0,\infty)$, what are the practical considerations I have to take care of? Obviously, the second derivative is not ensured to exists. Something more?

If $b$ is integral, then everything is fine; $f$ is an entire function, and the usual power series for the exponential will converge everywhere.

If $b$ is not integral, then there is a branch cut at $x = 0$, and the function is not analytic there. It is analytic on the open ray $(0,\infty)$, as you can easily check, but a power series centred at $r \in (0,\infty)$ will only have a radius of convergence equal to $r$.

Edit:

The 'side questions' are really where we start to see the difference between functions defined on the real line, and functions defined on the complex plane. Suppose $b$ is non-integral, and $b>1$. Then the first derivative can be defined on $[0,\infty)$, by the condition $f(x+\epsilon) - \epsilon f'(0) = \mathcal{O}(\epsilon^2)$ for $\epsilon \to 0^+$ (or $f'(0) := \lim_{x\to 0^+} f'(x)$). We could define a function $g$ on $(-\infty,0]$, for which $g(0) = f(0)$ and $g'(0^-) = f'(0^+)$. 'Gluing' $f$ and $g$ at zero gives a function which is continuously differentiable on $(-\infty,\infty)$. But, because $f$ has a branch cut at $x=0$, there is no way to extend this to a function which is complex differentiable at $0$.

Indeed, the special thing about holomorphic functions is the following: if $f(z)$ is continuously differentiable in some neighbourhood of $z_0$, then $f(z)$ is analytic at $z_0$.

I know the above is not a direct answer to side questions 1 and 2, but I think it gets to the relevant points.

• Thanks for the answer. Could you elaborate on the two side questions? That would be really helpful. Especially the the first one is interesting for me. – halirutan Sep 10 '13 at 13:41
• Thank you very much for sharing your insight. – halirutan Sep 10 '13 at 16:22