Show that $F_s(x)= |x|^{s-1}x$ is not smooth for all $s>1$ and $\frac{s-1}{2} \notin \mathbb N$

Asked Oct 06 '16 at 18:56

Active Oct 06 '16 at 19:13

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I am trying to prove that Show that $F_s(x)= ||x||^{s-1}x,x\in \mathbb R^n$ is not smooth for all $s>1$.

My strategy is as follows: To simplify the problem, I would only look at its partial derivative $D_1F_s(x)$ I get that $D_1F(x_1,\cdots,x_n) = (s-1)(x_1^2+\cdots+x_n^2)^{\frac{s-3}{2}}x_1 + ||x||^{s-1}$, then I argue that $(s-1)(x_1^2+\cdots+x_n^2)^{\frac{s-3}{2}}x_1$ is not continuous at $0$ if $s < 3$, and $\frac{s-1}{2} \notin \mathbb N$. For $5 > s \geqslant 3,\frac{s-1}{2} \notin \mathbb N$. just consider $D_1D_1f$. For $s$ larger just consider higher order derivatives.

However I am aware that the standard approach for this kind of problem is to use induction, but it would be troublesome to write down a formula for $D_1\cdots D_1f$, so how should I proceed if I am to use the induction?

edited Oct 06 '16 at 19:13

asked Oct 06 '16 at 18:56

Keith

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it is smooth for $\frac{s-1}2 \in \mathbb{N}$ since it is polynomial in $x_1,\ldots,x_n$ – reuns Oct 06 '16 at 19:04
Sorry about that, I have edited the question. – Keith Oct 06 '16 at 19:14
It is enough to look at $F_s(x_1,0,\ldots,0) = x_1 |x_1|^{s-1}$ – reuns Oct 06 '16 at 19:17
why would this be enough? – Keith Oct 06 '16 at 19:18
for what $s$ is it smooth ? – reuns Oct 06 '16 at 19:25
For $s$ odd it is smooth – Keith Oct 06 '16 at 19:34
let $g(t) = t |t|^{s-1}$ then $g'(t) = |t|^{s-1} + t (s-1)|t|^{s-2}sign(t) = s |t|^{s-1}$. $g'(t)$ is even, and for $t > 0$ : $g^{(k+1)}(t) = \frac{\Gamma(s)}{\Gamma(s-k)} t^{s-1-k}$ so that for every $t$ : $g^{(k+1)}(t) = \frac{\Gamma(s)}{\Gamma(s-k)} |t|^{s-1-k} (sign(t))^k$ that is smooth iff $\frac{s-1}{2} \in \mathbb{N}$ – reuns Oct 06 '16 at 19:54
excellent! Can you copy your comments into an answer so that I can accept it. – Keith Oct 06 '16 at 20:05

Show that $F_s(x)= |x|^{s-1}x$ is not smooth for all $s>1$ and $\frac{s-1}{2} \notin \mathbb N$

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