Let the function: \begin{equation} \phi(x)=\begin{cases} e^{-\frac{1}{1-x^2}}, & x\in (-1,1)\\ 0, &x \not \in (-1,1)\end{cases} \end{equation} How can I show that $\phi(x)$ is differentiable and that $\phi'(\pm1)=0$?
My approach
One similar problem that I have read is to prove that for the function $\alpha(x)=e^{-1/x}$ which considers the power series for the exponential function from which it holds that $\forall n \in \mathbb{Z}$, $e^y\geq \frac{y^n}{n!}, \forall y \geq 0$. Taking that into account, one can show that for the function $\alpha(x)=e^{-\frac{1}{x}}$ it holds: \begin{equation} \alpha(x)=\frac{1}{e^{\frac{1}{x}}} \leq \frac{n!}{\left(\frac{1}{x}\right)^n}=n!x^n \end{equation} which tells us that $\alpha(x)$ is differentiable at $0$ with $\alpha'(0)=0$.
- Why is this true?
- How can I apply this to my current problem?
I also tried to apply the definition of the derivative but it seems to be quite a complicated limit to solve.
Thank you!
We don't need exact formulas the derivatives of $\phi.$ In $(-1,1),$ every derivative of $\phi$ has the form $$e^{-1/(1-x^2)}p(x)(1-x^2)^{-m}$$ for some $m\in \{0,1,\dots \},$ where $p$ is a polynomial. Functions of this type $\to 0$ at $\pm 1$ from within $(-1,1).$ This implies $\phi^{(k)}(x) = 0, |x|\ge 1,$ for all $k,$ proving $\phi \in C^\infty(\mathbb {R}).$