$e^{-\frac{1}{x^2+y^2}}$ is $C^\infty$ and Taylor polynomial.

Question

Suppose $f\colon \mathbb{R}^2 \to \mathbb{R}$ is given by $$ f(x,y) = \begin{cases} e^{-\frac{1}{x^2+y^2}} & \textrm{ for } (x,y) \neq (0,0), \\ 0 & \textrm{ for } (x,y) = (0,0). \end{cases} $$ I'm trying to prove that it is a $C^\infty$-function on $\mathbb{R}^2$, and need to calculate the Taylorpolynomial of $f$ of order $n$.

I'm already stuck showing that it is $C^\infty$. My approach is to show that all the partial derivatives exist and are continuous. What I computed for the partial derivatives is the following: $$ D^k_1D^l_2f(x,y) = \frac{2^{k+l} x^ky^l}{(x^2+y^2)^{2k+2l}} e^{-\frac{1}{x^2+y^2}} $$ Therese are continuous outside of $(0,0)$, and for $(0,0)$ I am trying to show that $\lim_{(x,y) \to (0,0)} D^k_1D^l_2f(x,y) = 0$, but without any success. I also tried polar coordinates but I'm not sure if that works. Anyone knows how to proceed?

Answer 1

The situation is essentially the same but with less annoying notation in one dimension. Take $f(x)=\begin{cases} e^{-1/x^2} & x \neq 0 \\ 0 & x=0 \end{cases}$. Now you have the observations:

• If $x \neq 0$ then $f^{(n)}(x)=r_n(x) f(x)$ where $r_n(x)$ is a rational function whose denominator vanishes only at $x=0$.
• You evaluate $f^{(n)}(0)$ using the limit definition and show that it is zero.

To prove these you will want the observation:

• $p(x) e^{-x} \to 0$ as $x \to \infty$ for any polynomial $p$.

To extend to your situation you can just compose with $(x,y) \mapsto x^2+y^2$ which is easily seen to be $C^\infty$.

Answer 2

You will save some effort if you express $f$ as $g\circ h$, with$$\begin{array}{rccc}g\colon&\mathbb{R}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}e^{-\frac1{x^2}}&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}\end{array}$$and$$\begin{array}{rccc}h\colon&\mathbb{R}^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&x^2+y^2\end{array}.$$CLearly, $h$ is a $C^\infty$ functions and thereofre if you prove that $g$ is $C^\infty$ too, then $f$ will be $C^\infty$. Now, prove that $(\forall n\in\mathbb{N}):\lim_{x\to0}g^{(n)}(x)=0$. It will follow from this that $f$ is indeed $C^\infty$ and that its Taylor polynomial of any order is the null polynomial.