# Find a cut-off function in a ball.

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,

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

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

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?

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