Hoi, I want to show that for $\phi\in C_0^{\infty}(\mathbb{R}^n)$ where supp $\phi = \overline{B}(0,r):=B$
we have $$\sup_{x\in B}|\phi(x)|\leq 2R \sup|\partial_{x_1}\phi(x)| $$
I dont quite understand...
We can write $\phi(x) = \int_0^{x}(\partial_{x_1}\phi) dx_1$...right?
But then I get an estimation: $$\sup_{x\in B}|\phi(x)|\leq R \sup|\partial_{x_1}\phi(x)| $$
So how do we get the 2...
For every real numbers $x_2,\dots,x_n$, you have : $$||(-r,x_2,\dots,x_n)||=\sqrt{r^2+|x_2|^2+\dots+|x_n|^2}\geq r$$
Since supp $\phi= \overline{B}(0,r)=B$, you know that $\phi(x)=0$ whenever $||x||\geq r$.
This way you can write, for every $x\in B$ : $$\phi(x)=\phi(x)-\phi(-r,x_2,\dots,x_n)=\displaystyle\int_{-r}^{x_1} \partial_{x_1}\phi(t,\dots,x_n)dt$$
So you get, for all $x\in B$, $$|\phi(x)|\leq (x_1+r)\sup_{t\in[-r,x_1]} |\partial_{x_1} \phi(t,x_2,\dots,x_n)|\leq 2r \sup_{x\in B} |\partial_{x_1} \phi(x)|$$
Finally, you have your inequality :
$$\sup_B |\phi|\leq 2r \sup_B |\partial_{x_1}\phi|$$
