# Composition of sobolev functions

The following materials are exercise 8.10 and 8.11 from the functional analysis book by Prof. Brezis. The questions arise from whether we can well define a composition of two $W^{1,p}$ functions or not. From Corollary 8.11 we know if $G \in C^1, u \in W^{1,p}$, $G(u) \in W^{1,p}$ is well defined. However bad things might happen if $G \in W^{1,p}$. For instance, $G(x) = |x|$, then $G(u)$ might be bad on the set $\{ u = 0\}$. I thought we could still define things in the sense of distribution, but Prof. Brezis said in this special case we could have something more than that, because of the following propositions. The proof is not hard by following the hints, but the idea is really clever (IMO)!

Proposition 1 Let ${I=(0,1)}$ and ${u \in W^{1,p} (I)}$ with ${1 \le p < \infty}$. Then ${u' = 0}$ a.e. on the set ${E = \{ x \in I ; u(x) = 0\}}$.

Proof: Fix a function ${G \in C^1 ({\mathbb R} ;{\mathbb R})}$ such that ${|G(t)| \le 1, |G'(t)| \le C, \forall t \in {\mathbb R}}$, for some constant ${C}$, and

$\displaystyle G(t) = \begin{cases} 1 &\text{if } t\ge 1,\\ t &\text{if } |t| \le 1/2, \\ -1 & \text{if } t \le -1. \end{cases}$

Set ${v_n(x) = \frac{1}{n} G(n u(x)).}$

Since ${|v_n| \le \frac{1}{n}}$, ${v_n}$ converges to 0 uniformly.By Corollary 8.11(differentiation of a composition) ${v'_n = G'(n u(x)) u'(x)}$, we then have ${|v'_n| \le |u'| \in L^p}$. Hence ${v_n \in W^{1,p}}$. Define

$\displaystyle f(x) = \begin{cases} u'(x) &x \in E,\\ 0 & x \notin E. \end{cases}$

We can see that ${v'_n(x) \rightarrow f(x) }$ as ${n \rightarrow \infty}$ a.e. on ${I}$ by the construction of ${G}$. And ${|f(x)| \le |u'(x)|}$, hence in ${L^p}$. By Lebesgue dominated convergence theorem we will have ${v'_n \rightarrow f }$ in ${L^p}$. According to the definition of weak derivative, we have

$\displaystyle \int_I v'_n \varphi = -\int_I v_n \varphi' \quad \forall \varphi \in C_c^1 (I).$

For LHS we have ${v'_n \rightarrow f }$ in ${L^p}$, for RHS ${v_n \rightarrow 0}$ uniformly. Passing the limit we will have

$\displaystyle \int_I f \varphi = 0 \quad \forall \varphi \in C_c^1 (I).$

Which implies ${f = 0}$ a.e. And hence ${u' =0 }$ a.e. on ${E}$. $\Box$

Proposition 2 Let ${F \in C({\mathbb R}; {\mathbb R})}$ and assume that ${F \in C^1 ({\mathbb R} \setminus \{0\})}$ with ${|F'(t)| \le C, \forall t \in {\mathbb R} \setminus \{0\}}$, for some constant ${C}$. Let ${1 \le p < \infty}$. Then for every ${u \in W^{1,p}(0,1)}$ , we will have ${v = F(u) \in W^{1,p}(0,1)}$ and

$\displaystyle v'(x) = \begin{cases} F'(u(x))u'(x) &x \notin E,\\ 0 & x \in E. \end{cases}$

Furthermore, if ${\{u_k\}}$ a sequence in ${W^{1,p}(0,1)}$ such that ${u_k \rightarrow u}$ in ${W^{1,p}(0,1)}$, then ${F(u_k) \rightarrow F(u)}$ in ${W^{1,p}(0,1)}$.

Proof: Let ${\rho_n}$ be a sequence of mollifiers, define ${F_n = \rho_n \ast F}$, then ${F_n \in C^\infty}$, and ${F'_n = (\rho_n \ast F)' = \rho_n \ast F'}$. Hence ${\| F'_n \|_{L^\infty} \le \| \rho_n\| _{L^1} \|F'\|_{L^\infty} \le C}$. Denote ${v_n = F_n(u)}$, by Corollary 8.11 again, we have ${v_n \in W^{1,p}}$ and ${v_n' = F'_n(u) u'}$. Since ${u \in W^{1,p}}$, ${u \in L^\infty}$, ${v_n = F_n(u) \rightarrow v = F(u)}$ uniformly, and ${v_n' \rightarrow F'(u)u'}$ pointwise on the set where ${u(x) \neq 0}$. Define

$\displaystyle f(x) = \begin{cases} F'(u(x))u'(x) &x \notin E,\\ 0 & x \in E. \end{cases}$

Then

$\displaystyle \|v' - f\|_{L^p} \le \left( \int_{[0,1] \cap E^c} \left| F_n'(u)u' - F'(u)u' \right|^p\, dx \right)^ \frac{1}{p} + \left( \int_{[0,1] \cap E} \left| F_n'(u)u' \right|^p\, dx \right)^ \frac{1}{p}$

The first term goes to 0 by Lebesgue dominated convergence theorem since we have a uniform bound for ${F_n'}$ and ${F'}$, while the second term is identically zero by the previous proposition. Therefore ${v_n' \rightarrow f}$ in ${L^p}$. By the definition of weak derivative, passing the limit again we will have

$\displaystyle \int_I f \varphi = -\int_I v \varphi' \quad \forall \varphi \in C_c^1 (I).$

This implies ${v' = f}$ and ${v = F(u) \in W^{1,p}}$.

Now assume we have a sequence ${\{u_k\}}$ in ${W^{1,p}}$ and converges to ${u}$ in ${W^{1,p}}$, we want to show ${F(u_k) \rightarrow F(u)}$ in ${W^{1,p}}$.

Assume ${F(u_k)}$ does not converges to ${F(u)}$ in ${W^{1,p}}$, which means there exist an ${\varepsilon >0}$ and a subsequence ${\{u_{k_j}\}}$ such that ${\| u_{k_j} - u\|_{W^{1,p}} > \varepsilon, \forall k_j}$. Since ${u_{k_j} \rightarrow u}$ in ${W^{1,p}}$, there exists a sub-subsequence ${\{u_{k_{j_l}}\}}$ such that ${u_{k_{j_l}} \rightarrow u}$ a.e. with ${|u_{k_{j_l}}| \le h \in L^p}$, ${u_{k_{j_l}}' \rightarrow u'}$ a.e. with ${|u_{k_{j_l}}'| \le g \in L^p}$. It is easy to verify that ${F(u_{k_{j_l}}) \rightarrow F(u)}$ in ${W^{1,p}}$, which leads to the contradiction. $\Box$