3
$\begingroup$

Let $0\le r< R\le 1$. How do we find a function $\eta\in C^1(\mathbb{R})$ such that $\eta=1$ in $B_r$ (the ball center at $0$ and radius $r$) and $\eta=0$ outside $B_R$ and $|D\eta|\le \frac{2}{R-r}$ ?

I tried to convolve the function ($\phi$ is a mollifier) $$\left(\frac{4}{R-r}\right)^n\phi\left(\frac{4x}{R-r}\right),$$

with the characteristic function of the ball $B_{(R+r)/3}$, however, I got no success.


The function $f$ equals to zero when $|x|\ge R-\epsilon$ and it equals to $1$ when $|x|\le R+\epsilon$. Therefore, we have $$\eta(x)=\int_{B_{R-\epsilon}}\phi_{\delta}(x-y)f(y)dy.$$ Notice that $\phi_{\delta}(z)=0$ when $|z|>\delta$. When $|x|\ge R$, we have $|x-y|\ge \epsilon>\delta$ for all $y\in B_{R-\epsilon}$.Therefore, it is clearly that $\eta(x)=0$. when $|x|<r$, for all $y$ such that $|y|>r+\epsilon$, we also have $|x-y|>\delta$. Therefore, we have $$\eta(x)=\int_{B_{r+\epsilon}}\phi_{\delta}(x-y)f(y)dy=\int_{B_{r+\epsilon}}\phi(x-y)dy.$$

Now, we have $B_{\delta}(x)\subset B_{r+\epsilon}$, so $$\eta(x)=\int_{B_{r+\epsilon}}\phi(x-y)dy=\int_{\mathbb{R}^n}\phi_{\delta}(z)dz=1.$$
To conclude, notice that with $\epsilon=\frac{R-r}{4}$, we have $|f'|\le \frac{2}{R-r}$ a.e. Therefore,

$$|\eta'(x)|\le\frac{2}{R-r}\int_{\mathbb{R}^n}\phi_{\delta}(x-y)dy=\frac{2}{R-r}$$

$\endgroup$
3
  • $\begingroup$ Have you tried something? $\endgroup$ – Tomás May 26 '15 at 13:09
  • $\begingroup$ I used convolution. But I can not estimate the derivative. I used the mollifier function $\eta$. Then I set $\eta^{\frac{R-r}{4}}$ $\endgroup$ – Omega May 26 '15 at 13:36
  • $\begingroup$ I used the mollifier function $\eta$. Then I set $\eta^{\frac{R-r}{4}}(x)=(\frac{4}{R-r})^n\eta(\frac{4x}{R-r})$. Then I convoluted $\eta^{\frac{R-r}{4}}$ with $\chi_{B_{\frac{R+r}{2}}}$. $\endgroup$ – Omega May 26 '15 at 13:46
1
$\begingroup$

Choose $\epsilon>0$ such that $$\epsilon\le\frac{R-r}{4}.$$

Consider the function $f:\mathbb{R}\to \mathbb{R}$ defined by $$f(x)=\max\left\{0,\min\left\{1,\frac{R-\epsilon-|x|}{R-r-2\epsilon}\right\}\right\}.$$

Now, consider a mollifier $\phi$ and set $\phi_\delta(x)=\delta^{-1}\phi(x/\delta)$. For $\delta<\epsilon$, consider the function $$\eta(x)=(\phi_\delta\star f)(x),\ x\in \mathbb{R}.$$

Can you prove that $\eta$ is your desired function?

$\endgroup$
10
  • $\begingroup$ i am sorry but I have no idea to prove it. Could you please give me some hints ? Thank you $\endgroup$ – Omega May 26 '15 at 14:41
  • $\begingroup$ Well, try it first. If you not try, you will not understand it. I suggest you to draw a picture of $f$ and understand why I choose $\epsilon$ and $\delta$ as I did. $\endgroup$ – Tomás May 26 '15 at 14:43
  • $\begingroup$ Ok. I'll try now but I think it will take time. Are you still here tomorow ? $\endgroup$ – Omega May 26 '15 at 14:51
  • $\begingroup$ Yes, I will be here. Take your time and try to understand it first, then you come back here. $\endgroup$ – Tomás May 26 '15 at 15:03
  • $\begingroup$ Hello, I wrote my answer. I still can not estimate the derivative. I tried to calculate the convolution but I still did not success. $\endgroup$ – Omega May 27 '15 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.