Let $I=(0,1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.9
- Let $u \in W^{2, p}(I)$ with $1<p<\infty$. Assume that $u(0)=u^{\prime}(0)=0$. Show that $\frac{u(x)}{x^2} \in L^p(I)$ and $\frac{u^{\prime}(x)}{x} \in L^p(I)$ with (1) $$ \left\|\frac{u(x)}{x^2}\right\|_{L^p(I)}+\left\|\frac{u^{\prime}(x)}{x}\right\|_{L^p(I)} \leq C_p\left\|u^{\prime \prime}\right\|_{L^p(I)} $$
- Deduce that $v(x)=\frac{u(x)}{x} \in W^{1, p}(I)$ with $v(0)=0$.
- Let $u$ be as in question $(1)$. Set $u_n=\zeta_n u$, where $\zeta_n$ is defined in question 4 of Exercise 8.8. Check that $u_n \in W^{2, p}(I)$ and $u_n \rightarrow u$ in $W^{2, p}(I)$ as $n \rightarrow \infty$.
- More generally, let $m \geq 1$ be an integer, and let $1<p<\infty$. Assume that $u \in X_m$, where $$ X_m=\left\{u \in W^{m, p}(I) : u(0)=D u(0)=\cdots=D^{m-1} u(0)=0\right\} . $$ Show that $\frac{u(x)}{x^m} \in L^p(I)$ and that $\frac{u(x)}{x^{m-1}} \in X_1$.
- Assume that $u \in X_m$ and prove that $$ v=\frac{D^j u(x)}{x^{m-j-k}} \in X_k \quad \forall j, k \text { integers, } j \geq 0, k \geq 1, j+k \leq m-1. $$
- Let $u \in X_m$ and $\zeta \in C^\infty (\mathbb R)$ such that $\zeta (x)=0$ for $x \le 1$ and $\zeta (x)=1$ for $x \ge 2$. Let $\zeta_n (x) := \xi (nx)$ and $u_n (x) = \zeta_n (x) u(x)$. Prove that $\zeta_n u \in W^{m, p}(I)$ and $\zeta_n u \rightarrow u$ in $W^{m, p}(I)$, as $n \rightarrow \infty$.
There are possibly subtle mistakes that I could not recognize in my below attempt of $(6)$. Could you please have a check on it?
We will proceed by induction on $m \ge 1$. The base case $m=1$ has been proved in Exercise 8.8.4. We assume that for $1 \le k \le m$ that $u \in X_k$ implies $u_n :=\zeta_n u \in W^{k, p}(I)$ and $u_n \xrightarrow{n \to \infty} u$ in $W^{k, p}(I)$. We will prove that the statement holds for $k=m+1$. Fix $u \in X_{m+1}$.
By general Leibniz rule, $$ \begin{align*} D^k u_n = D^k (\zeta_n u) = \sum_{i=0}^k {k \choose i} (D^i \zeta_n) (D^{k-i}u), \quad k =0, \ldots, m+1. \end{align*} $$
We have $D^i \zeta_n \in L^\infty (I)$ and $D^i u \in L^p (I)$ for all $i =0, \ldots, m+1$. Then $D^k u_n \in L^p (I)$ for all $k =0, \ldots, m+1$. Then $u_n \in W^{m+1, p}(I)$.
Next we prove $u_n \xrightarrow{n \to \infty} u$ in $W^{m+1, p}(I)$. It follows from $u \in W^{m+1, p}(I)$ that $u, u' \in W^{m, p}(I)$. By inductive hypothesis, $u_n \to u$ in $W^{m, p}(I)$. It remains to prove $D^{m+1} u_n \xrightarrow{n \to \infty} D^{m+1} u$ in $L^{p}(I)$. Notice that $$ D^{m+1} u_n = \zeta_n (D^{m+1}u) + \sum_{i=1}^{m+1} {m+1 \choose i} (D^i \zeta_n) (D^{m+1-i}u) $$ and that $$ D^i \zeta_n \xrightarrow{n \to \infty} 0 \quad \text{everywhere}, \quad i =1, \ldots, m+1. $$
I already proved that $u(0)$ implies $\zeta_n' u \to 0$ in $L^{p}(I)$. Notice that $u \in X_{m+1}$ implies $u(0)=Du(0) = D^2u(0) = D^mu=0$. With similar reasoning, we can prove for $i =1, \ldots, m+1$ that $(D^i \zeta_n) (D^{m+1-i}u) \to 0$ in $L^{p}(I)$. The claim then follows by the fact that $\zeta_n (D^{m+1}u) \xrightarrow{n \to \infty} D^{m+1}u$ in $L^{p}(I)$.
Fix $i \in \{ 1, \ldots, m+1 \}$ and let $j := m+1-i$. I think the proof of $(D^i \zeta_n) (D^j u) \to 0$ in $L^{p}(I)$ needs more details.
Notice that $\operatorname{supp} (D^i \zeta) \subset \operatorname{supp} \zeta = [\frac{1}{n}, \frac{2}{n}]$ and $i=m+1-j$. Then $$ \begin{align*} I_n &:= \int_0^1 | (D^i \zeta_n) (x) (D^j u) (x) |^p dx \\ &= \int_{1/n}^{2/n} | n^i D^i \zeta (nx) (D^j u) (x) |^p dx \\ &= \int_{1/n}^{2/n} | (nx)^i D^i \zeta (nx) |^p \left | \frac{(D^j u) (x)}{x^{i}} \right |^p dx \\ &\le \| 2^i D^i \zeta \|^p_\infty \int_{1/n}^{2/n} \left | \frac{(D^j u) (x)}{x^{i}} \right |^p dx \\ &= \| 2^i D^i \zeta \|^p_\infty \left \| \frac{(D^j u) (x)}{x^{i}} \right \|^p_{L^p (\frac{1}{n}, \frac{2}{n})}. \end{align*} $$
Notice that $u \in X_{m+1}$ implies $D^j u \in X_{m+1-j} = X_i$. By $(4)$, $$ \frac{(D^j u) (x)}{x^{i}} \in L^p (I). $$
Then $$ \left \| \frac{(D^j u) (x)}{x^{i}} \right \|^p_{L^p (\frac{1}{n}, \frac{2}{n})} \xrightarrow{n \to \infty} 0. $$
This completes the proof.