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For any $R>0,$ I need an example of a family of functions $\phi_R\in C_c^{1}(\mathbb{R}^N)$ such that $0\leq \phi_R\leq 1$ in $\mathbb{R}^N$ satifying $\phi_R=1$ in $B(0,R)$ and $\phi_R=0$ in $\mathbb{R}^N\setminus B(0,2R)$ such that $$ |\nabla\phi_R|\leq \frac{c}{R} $$ for some positive constant $c$ independent of $R$.

I tried in the following way:

Let $\phi\in C_c^{1}{\mathbb{R}^N}$ such that $0\leq \phi\leq 1$ in $\mathbb{R}^N$ and $\phi(t)=1$ if $|t|\leq 1$ and $\phi(t)=0$ if $|t|\geq 2$. Then define for any $R>0$ $$ \phi_R=\phi(\frac{|x|}{R}) $$ Then it follows from the property of $\phi$ that $0\leq \phi_R\leq 1$ in $\mathbb{R}^N$ together with $\phi_R=1$ in $B(0,R)$ and $\phi_R=0$ in $\mathbb{R}^N\setminus B(0,2R)$. But I am unable to prove the last condition $$ |\nabla\phi_R|\leq \frac{c}{R} $$ for $c$ independent of $R$.

Please help me.

Thank you.

Mathguide
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  • I think that the standard construction is to convolute a bump function with a characteristic function, see wiki. Though you might also be able to get your final inequality using the chain rule. – MSDG Jul 06 '18 at 09:49
  • A concrete counterexample from your side can be much more helpful for me. Thanks – Mathguide Jul 06 '18 at 10:16
  • I did not claim that your construction does not work, I only gave you the reference that I know of. – MSDG Jul 06 '18 at 11:15
  • I have tried your example, but I am again unable to derive the gradiant estimate, which is my main question. Please help. – Mathguide Jul 06 '18 at 13:29

2 Answers2

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In your example, notice first of all that $\phi_R$ is differentiable, since it is equal to 1 on a neighbourhood of the origin, so the absolute value in the argument poses no problem.

Furthermore, $\phi_R$ vanishes outside of the ball $B(0,2R)$, so our bound really only depends on the values of $\phi_R$ inside the ball.

The chain rule gives $$|\partial_j \phi_R(x)| = |\partial_j \big(\phi\left(|x|/R\right)| \big) = |\partial_j \phi(|x|/R)| \cdot \frac{1}{R}. $$ where the equalities follow since we are taking the absolute value anyway (otherwise there could be problems with the signs).

Now since $\phi \in C^1_c$, all of its partial derivatives are in particular continuous, and hence bounded on the compact set $B(0,2R)$, so that $|\partial_j \phi(|x|/R)| \leq C_j$ for each $j = 1, \ldots, N$. Hence, $$ |\nabla \phi_R| \leq \sum_{j=1}^N |\partial_j \phi_R| \leq \sum_{j=1}^N \frac{C_j}{R} = \frac{C}{R}, $$ with $C = \sum_{j=1}^N C_j.$

MSDG
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Choose $\epsilon>0$, and use the construction given here and use $$\phi(x)= f\left(\frac{|x|}{R}\right).$$ The gradient is bounded by $\frac{1+\epsilon}R$.

username
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