Let $f: \mathbb{R}^n\rightarrow \mathbb{R}$ be the following function, $$f(x)=\begin{cases} \operatorname{e}^{-\tfrac{1}{1-\|x\|^2}} & \text{if }\|x\|<1,\\\\ 0 & \text{otherwise}. \end{cases}$$ How can I show that $f$ is smooth?
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Of course $x\longmapsto 1- \|x\|^2$ is smooth. By composition, it is enough to show that the real function defined by $h(t)=e^{-1/t}$ for $t>0$ and $h(0)=0$ for $t\leq 0$ is smooth. The only problem is at $0$. Prove by induction that for every $t>0$ $$ h^{(n)}(t)=\frac{p_n(t)}{t^{2n}}e^{-1/t} $$ with $p_n$ a polynomial. It follows that for every $n\geq 0$ $$ \lim_{t\rightarrow 0^+}h^{(n)}(t)=0. $$ So $h$ is infinitely many times differentiable at $0$, with $h^{(n)}(0)=0$ for every $n\geq 0$.
Note: the composition is $f(x)=h(1-\|x\|^2)$.
Julien
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When you say "By composition", do you mean the composition of $f(t)$ in your answer and $g(x)=1-||x||^2$? but $g$ is not differentiable at 0, what am I missing? – user70694 Apr 03 '13 at 00:01
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@user70694 Exactly. And yes, $g$ is a polynomial in the coordinates. It is smooth. It is $|x|$ which is not differentiable at $0$. Check the square! – Julien Apr 03 '13 at 00:02
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@user70694 Do you agree now that $1-|x|^2=1-x_1^2-\ldots-x_n^2$ is smooth? And do you need more details in the answer? – Julien Apr 03 '13 at 00:41
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Define $g:\mathbb{R}^{\ge0}\mapsto\mathbb{R}$ by $$ g(t)=\left\{\begin{array}{} e^{\frac1{1-t^2}}&\text{for }0\le t\lt1\\ 0&\text{for }t\ge1 \end{array}\right. $$
At all points other than $t=1$, the function is a composition of smooth functions. So we only need to show that all the derivatives are $0$ when $t=1$.
For $t\ge1$, $g$ is identically $0$, so there is no problem there.
For $0\le t\lt1$, induction gives us $\frac{\mathrm{d}^n}{\mathrm{d}t^n}g(t)=R_n(t)g(t)$ for some rational function $R_n$. Since $\lim\limits_{x\to\infty}P(x)e^{-x}=0$ for any polynomial $P$, we have that $\lim\limits_{t\to1^-}R_n(x)g(x)=0$.
Since all derivatives tend to $0$ as $t\to1$, the mean value theorem says that all the derivatives of $g(t)$ at $t=1$ are $0$.
Since $x\mapsto|x|$ is smooth away from $0$, we have that $$ f(x)=g(|x|)=e^{\frac1{1-|x|^2}} $$ is smooth away from $0$. Near $0$, $f$ is the composition of smooth functions. Thus, $f$ is smooth over all of $\mathbb{R}^n$.
robjohn
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Are you implicitly bringing this back to $\mathbb{R}$ like me? The function is defined on $\mathbb{R}^n$. – Julien Apr 03 '13 at 00:01
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@julien: I had to break for dinner and walking the dog, but now I'm not :-) – robjohn Apr 03 '13 at 01:46
$$\nabla f = \frac{-2xf}{(1-|x|^2)^2}$$
– Fly by Night Apr 02 '13 at 23:31