Proving that $\lambda$ is not analytic in differential topology.

Question

Let $\lambda : \mathbb{R} \to \mathbb{R}$ be defined by

$\lambda(t) =\begin{cases} 0, & \text{for } t \leq 0 \\ e^{-1/t}, & \text{for } t > 0. \end{cases}$

This is a smooth function with values between zero and one. Note that all derivatives at zero are zero: the McLaurin series fails miserably and $\lambda$ is definitely not analytic.

My question is why the McLaurin series fails miserably? and how this leads to that $\lambda$ is not anaytic? and what does it mean for $\lambda$ to be analytic in this context? Is the notion of analytic function in differential topology different from its meaning in complex analysis?

Answer 1

For "analytic", you should read "can be written as a power-series". This is equivalent to being holomorphic (complex-differentiable on an open set) in a complex analysis setting, but is also a sensible definition when you're doing real analysis (or differential topology/geometry).

The Maclaurin series "fails miserably" because all the function's derivatives are zero. Thus the Maclaurin series of the function is the zero series, so it definitely does not converge to $\lambda$. If $\lambda$ has a power series expansion at $x=0$, then that power series must be the Maclaurin series of $\lambda$. So $\lambda$ is not analytic (has no power series expansion) at $x=0$.

Answer 2

If $\lambda$ were to be real-analytic, then there would be $\epsilon>0$ such that $$\lambda(t) = \sum_{k\geq 0} \frac{\lambda^{(k)}(0)}{k!}t^k$$whenever $|t|<\epsilon$, with uniform convergence of that series. But if all derivatives vanish, the series is the zero series. And $\lambda(t)>0$ whenever $t>0$, which would be a contradiction. Being real-analytic is stronger than being $C^\infty$ and this $\lambda$ is the classical example.