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Show that $f:\Bbb R^m\to\Bbb R,\: x\mapsto [|x|<1] \exp\left(\frac1{|x|^2-1}\right)$ is smooth, where $[|x|<1]$ is a Iverson bracket.

My work so far: $f|_{\{x\in\Bbb R^m:|x|<1\}}$ is a composition of smooth functions, hence by the chain rule it can be shown that it is also smooth, and $f|_{\{x\in\Bbb R^m:|x|\ge 1\}}=0$ is trivially smooth. Then we must check that $f$ is differentiable in $\mathrm S^{m-1}$, that is, we must show that

$$\lim_{h\to 0}\frac{\|\partial^k f(x+h)-\partial^k f(x)-\partial^{k+1}(x)h\|}{\|h\|}=0,\quad\forall k\in\Bbb N_{>0},\:\forall x\in\mathrm S^{m-1}\tag1$$

I guess that $\partial^k f(x)=0$ for all $x\in\mathrm S^{m-1}$ and for all $k\in\Bbb N$. Then, assuming this hypothesis, from $(1)$ the proof was reduced to show that

$$\partial^k f(x+h)\in o(\|h\|)\quad\text{for }|x+h|<1,\:x\in\mathrm S^{m-1}\tag2$$

Because $f$ is smooth in this region using the Taylor theorem we have that

$$f(x+h)=\sum_{k=0}^n\frac{\partial^kf(h)[x]^k}{k!}+o(\|x\|^n),\quad\forall n\in\Bbb N\tag3$$

where $o(\|x\|^n)=o(1)$ because $x\in\mathrm S^{m-1}$. However I cant stablish a clear relation between $(2)$ and $(3)$ to end the proof. I need some help with this exercise.

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    $\begingroup$ I would suggest you first attempt this problem with $m = 1$. Then, try to carry that method into more dimensions. $\endgroup$ – астон вілла олоф мэллбэрг Sep 18 '17 at 5:04
  • $\begingroup$ @астонвіллаолофмэллбэрг thank you but I tried this approach before and it didnt work. I need to show it for $m=2$, with $m=1$ it is not enough. But for $m=2$ the mixed partial derivatives have a painful form, so I didnt get anything from here. $\endgroup$ – Masacroso Sep 18 '17 at 5:25
  • $\begingroup$ oh... wait, it can work using strong induction... Or maybe not... $\endgroup$ – Masacroso Sep 18 '17 at 5:26
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Prove the following Lemma using induction.

Lemma. The function $q(t):=e^{-1/t}$ $(t>0)$ and $:=0$ $(t\leq0)$ is smooth, whereby $q^{(n)}(0)=0$ for all $n\geq0$.

Your $f$ is just $q\circ g$, where $g(x):=1-|x|^2$. The chain rule then guarantees the claim in your question. When required do an additional induction proof for this. But it is not necessary to compute partial derivatives.

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  • $\begingroup$ I know the case for $q$ but I have trouble reducing $q\circ g$ to $q$. In first place it doesnt seems totally clear that the chain rule solves all... , that is $\partial(q\circ g)(x)=(\partial q\circ g)(x)\partial g(x)$, then the complicated part is justify that the derivatives of $g$ holds the original claim. $\endgroup$ – Masacroso Sep 18 '17 at 9:13
  • $\begingroup$ but I think I finally solved it just seeing that $$\|\partial^kg(x)\|\le\frac{C}{\||x|^2-1\|^{k+1}},\quad |x|<1$$ for some positive constant $C$. $\endgroup$ – Masacroso Sep 18 '17 at 9:17

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