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I am looking for a $C^\infty $ function $g:\mathbb{R}^3\to \mathbb{R}^3$ such that $g(x)=0$ for $|x|\le 1$ and $g(x)=x$ for $|x|\ge 2$. Certainly such $g$ can be constructed, but I also want it to satisfy the additional property that for each $j=1,2,3$,

$$\sum _{k=1}^3 \left |\frac{\partial ^2 g_k}{\partial x_j \partial x_k}\right | \le C\sum _{i=1}^3 \left | \frac{\partial g_i}{\partial x_i}\right |$$

for some constant $C>0$.

Can I find $g$?

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Could a $g$ of the form $g(x) = f(|x|)x$ work? –  Dirk Nov 24 '11 at 13:45
    
Hi Dirk, yes, any $g$ as described would work. –  flavio Nov 24 '11 at 13:48
    
Well, I meant: Did you try to construct an $f$ as above? This would be the direction I would take... –  Dirk Nov 24 '11 at 13:56
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Define $\phi(t) = \exp(-1/(1-t^2))$ (with zero extension), define $h(t) = c\int_{-\infty}^t \phi(s)ds$ (with appropriate c) and set $f(t) = h(x-2)$. Then define $g(x) = f(|x|^2)x$ and perform tedious calculations... –  Dirk Nov 24 '11 at 14:15
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Either I misread the problem, or the function $f=\sum_i|(g_i)_{x_i}|$ satisfies $|\nabla f|\le Cf$, so, by Gronwall, $f=0$, whence $g_1$ does not depend on $x_1$, etc. But we want $g_1=x_1$ for large $x$. –  fedja Nov 24 '11 at 14:21

1 Answer 1

@GH: Considering one variable, assume $|u'(t)|\le C|u(t)|$ and that $u(t)=0$ for $|t|\le 1$. Now $|u(t)|^2 = u(t)^2$ implies

$$2|u(t)|\frac{d|u(t)|}{dt} = 2u(t)u'(t)$$

whence

$$\frac{d|u(t)|}{dt} \le u'(t)\le C|u(t)|.$$

Now the function $f(t):=|u(t)|$ is non-negative, satisfies $f'(t)\le Cf(t)$. By Grönwall's inequality we obtain

$$f(t)\le f(0)e^{Ct}=0 \quad \text{for }t>0$$

and thus $u(t)=0$ for $t\ge 0$. The case $t<0$ can be treated similarly.

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Thank you Alex! –  GH from MO Dec 9 '11 at 15:28

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