2

Question:

I need to prove the following chain rule:

Let $F:\mathbb{R}\rightarrow\mathbb{R}$, $F\in C^1$ with $F'$ bounded.

Let $U$ bounded and $u\in W^{1,p}(U)$ with $1\leq p\leq\infty$. Show that $v:=F(u)\in W^{1,p}(U)$.

My solution: There exists $\epsilon>0$ such that

$F(t)-F(0)=F'(\epsilon)t$ (because $F\in C^1$)

$\Rightarrow |F(t)|=|F'(\epsilon)||t|+|F(0)|\leq M|t|+|F(0)|$ (because $F'$ is bounded)

$\Rightarrow |F(u(t))|\leq M|u(t)|+|F(0)|$

I need to show that

$\displaystyle\int_U|F(u)|^pdx<\infty$

but I don't know how to do it.

Thanks in advance for your help

Cookie
  • 13,532
yemino
  • 814

1 Answers1

2

Recall, that for $(x_1, x_2) \in \mathbb R^2$, we have $$ \|(x_1,x_2)\|_1 = |x_1| + |x_2| \le 2^{(p-1)/p} \, \|(x_1,x_2)\|_p = 2^{(p-1)/p} \, (|x_1|^p + |x_2|^p)^{1/p}. $$ Hence, $$|F(u)|^p \le \big| M \, |u| + |F(0)| \big|^p \le 2^{p-1} \, (M^p \, |u|^p + |F(0)|^p ).$$ This shows $\int |F(u)|^p dx < \infty$.

gerw
  • 31,359