I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.8.4 Let $I=(0,1)$ and $u \in W^{1, p} (I)$ with $p \in [1, \infty)$ and $u(0)=0$. Fix a function $\xi \in C^\infty (\mathbb R)$ such that $\xi (x)=0$ for $x \le 1$ and $\xi (x)=1$ for $x \ge 2$. Let $\xi_n (x) := \xi (nx)$ and $u_n (x) = \xi_n (x) u(x)$.
- Check that $u_n \in W^{1, p} (I)$.
- Prove that $u_n \to u$ in $W^{1, p} (I)$.
In my below attempt, I can prove $(2)$ for $p \in (1, \infty)$. Could you explain how to prove it for $p=1$?
First, we have $\| \xi_n \|_\infty = \| \xi \|_\infty$ for all $n$. Second, $\xi_n (x)=0$ for $x \le \frac{1}{n}$ and $\xi (x)=1$ for $x \ge \frac{2}{n}$. Then $\xi_n \to 1$ everywhere on $(0, 1)$ and $\|\xi_n-1 \|_q \to 0$ for $q \in [1, \infty)$.
Second, $\xi_n' (x) = n\xi' (nx)$. Then $\xi_n' (x)=0$ for $x < \frac{1}{n}$ and for $x > \frac{2}{n}$. Then $\xi'_n \to 0$ everywhere on $(0, 1)$.
Third, $u'_n (x) = \xi_n' (x) u(x) + \xi_n (x) u'(x)$. Clearly, $u'_n \to u'$ everywhere on $(0, 1)$.
1.
Clearly, $\xi_n \in W^{1, p} (I)$ for all $n$. Then $u_n \in W^{1, p} (I)$ by below result (in the same book), i.e.,
Corollary 8.10 Let $I$ be a (possibly unbounded) open interval of $\mathbb R$. Let $u, v \in W^{1, p}(I)$ with $1 \leq p \leq$ $\infty$. Then $u v \in W^{1, p}(I)$ and $(u v)^{\prime}=u^{\prime} v+u v^{\prime}$. Furthermore, the formula for integration by parts holds: $$ \int_y^x u^{\prime} v=u(x) v(x)-u(y) v(y)-\int_y^x u v^{\prime} \quad \forall x, y \in \bar{I} . $$
Clearly, $u_n \to u$ everywhere on $(0, 1)$. By dominated convergence theorem, $u_n \to u$ in $L^p$. It remains to prove that $u'_n \to u'$ in $L^p$. It suffices to prove $$ I_n = \| \xi_n' u \|_p^p = \int_0^1 |n\xi' (nx)u(x)|^p dx = \int_0^{2/n} |n\xi' (nx)u(x)|^p dx \to 0. $$
Then $$ \begin{align*} I_n &= \int_0^1 |n\xi' (nx)u(x)|^p dx \\ &= \int_0^{2/n} |nx\xi' (nx)|^p\left | \frac{u(x)}{x} \right |^p dx \\ &\le 2 \| \xi' \|_{L^\infty (I)} \int_0^{2/n} \left | \frac{u(x)}{x} \right |^p dx. \end{align*} $$
By Exercise 8.8.1 (in the same book), $u(0)=0$ and $p \in (1, \infty)$ implies $$ \left\|\frac{u(x)}{x}\right\|_{L^p(I)} \leq \frac{p}{p-1}\left\|u^{\prime}\right\|_{L^p(I)}, $$ and thus $$ \int_0^{2/n} \left | \frac{u(x)}{x} \right |^p dx \to 0. $$
Then it remains to prove for the case $p=1$.
Let $p=1$. Then
$$ \begin{align*} I_n &= \int_0^1 |n\xi' (nx)u(x)| dx \\ &= \int_0^{2/n} |n\xi' (nx)u(x)| dx \\ &\le n \| \xi' \|_{L^\infty (0, 2)} \int_0^{2/n} |u(x)| dx \\ &\le 2\| \xi' \|_{L^\infty (0, 2)} \sup_{x\in (0,\frac{2}{n})} |u(x)|. \end{align*} $$
Because $u \in C (\bar I)$ and $u(0)=0$, we get $$ \sup_{x\in (0,\frac{2}{n})} |u(x)| \to 0. $$
The claim then follows.