$C_c^\infty(\mathbf{R})$ is dense in $W^{1,p}(\mathbf{R})$

Question

I am dealing with this theorem from Brezis' book. I attach it and its proof below. At step c) the author considers the $L^p$-norm of $u_n-u$ and writes that inequality. But I cannot understand why it holds: I thought it could be true if $L^p$-norm of $\zeta_n$ was $1$ but, if I am not wrong, it depends on $n$ and grows with it.

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Relevant pages of the book (transcribed from [Image 1] [Image 2] [Image 3]):

Theorem 8.7 (density). Let $u \in W^{1, p}(I)$ with $1 \leq p<\infty .$ Then there exists $a$ sequence $\left(u_{n}\right)$ in $C_{c}^{\infty}(\mathbb{R})$ such that $u_{n \mid I} \rightarrow u$ in $W^{1, p}(I) .$

Remark 9. In general, there is no sequence $\left(u_{n}\right)$ in $C_{c}^{\infty}(I)$ such that $u_{n} \rightarrow u$ in $W^{1, p}(I)$ (see Section 8.3 ). This is in contrast to $L^{p}$ spaces: recall that for every function $u \in L^{p}(I)$ there is a sequence $\left(u_{n}\right)$ in $C_{c}^{\infty}(I)$ such that $u_{n} \rightarrow u$ in $L^{p}(I)$ (see Corollary 4.23 ).

Proof. We can always suppose $I=\mathbb{R} ;$ otherwise, extend $u$ to a function in $W^{1, p}(\mathbb{R})$ by Theorem $8.6 .$ We use the basic techniques of convolution (which makes functions $\left.C^{\infty}\right)$ and cut-off (which makes their support compact).

(a) Convolution.

We shall need the following lemma.

Lemma 8.4 . Let $\rho \in L^{1}(\mathbb{R})$ and $v \in W^{1, p}(\mathbb{R})$ with $1 \leq p \leq \infty .$ Then $\rho \star v \in$ $W^{1, p}(\mathbb{R})$ and $(\rho \star v)^{\prime}=\rho \star v^{\prime}.$

Proof. First, suppose that $\rho$ has compact support. We already know (Theorem 4.15) that $\rho \star v \in L^{p}(\mathbb{R}) .$ Let $\varphi \in C_{c}^{1}(\mathbb{R}) ;$ from Propositions 4.16 and 4.20 we have $$ \int(\rho \star v) \varphi^{\prime}=\int v\left(\check{\rho} \star \varphi^{\prime}\right)=\int v(\check{\rho} \star \varphi)^{\prime}=-\int v^{\prime}(\check{\rho} \star \varphi)=-\int\left(\rho \star v^{\prime}\right) \varphi, $$ from which it follows that $$ \rho \star v \in W^{1, p} \quad \text { and } \quad(\rho \star v)^{\prime}=\rho \star v^{\prime}. $$ If $\rho$ does not have compact support introduce a sequence $\left(\rho_{n}\right)$ from $C_{c}(\mathbb{R})$ such that $\rho_{n} \rightarrow \rho$ in $L^{1}(\mathbb{R})$ (see Corollary 4.23). From the above, we get $$ \rho_{n} \star v \in W^{1, p}(\mathbb{R}) \quad \text { and } \quad\left(\rho_{n} \star v\right)^{\prime}=\rho_{n} \star v^{\prime}. $$ But $\rho_{n} \star v \rightarrow \rho \star v$ in $L^{P}(\mathbb{R})$ and $\rho_{n} \star v^{\prime} \rightarrow \rho \star v^{\prime}$ in $L^{p}(\mathbb{R})$ (by Theorem 4.15). We conclude with the help of Remark 4 that $$ \rho \star v \in W^{1, p}(\mathbb{R}) \quad \text { and } \quad(\rho \star v)^{\prime}=\rho \star v^{\prime}. $$ (b) Cut-off.

Fix a function $\zeta \in C_{c}^{\infty}(\mathbb{R})$ such that $0 \leq \zeta \leq 1$ and $$ \zeta(x)=\left\{\begin{array}{ll} 1 & \text { if }|x|<1, \\ 0 & \text { if }|x| \geq 2. \end{array}\right. $$ Define the sequence (4) $$ \zeta_{n}(x)=\zeta(x / n) \quad \text { for } n=1,2, \ldots. $$ It follows easily from the dominated convergence theorem that if a function $f$ belongs to $L^{p}(\mathbb{R})$ with $1 \leq p<\infty,$ then $\zeta_{n} f \rightarrow f$ in $L^{p}(\mathbb{R})$.

(c) Conclusion.

$\bbox[lightgreen]{\text{Choose a sequence of mollifiers}\Rule{0px}{0.5em}{0.5em}} $$\left(\rho_{n}\right) .$ We claim that the sequence $u_{n}=\zeta_{n}\left(\rho_{n} \star u\right)$ converges to $u$ in $W^{1, p}(\mathbb{R}) .$ First, we have $\left\|u_{n}-u\right\|_{p} \rightarrow 0 .$ In fact, write $$ u_{n}-u=\zeta_{n}\left(\left(\rho_{n} \star u\right)-u\right)+\left(\zeta_{n} u-u\right) $$ and thus $$ \left\|u_{n}-u\right\|_{p} \leq\left\|\rho_{n} \star u-u\right\|_{p}+\left\|\zeta_{n} u-u\right\|_{p} \rightarrow 0. $$ Next, by Lemma $8.4,$ we have $$ u_{n}^{\prime}=\zeta_{n}^{\prime}\left(\rho_{n} \star u\right)+\zeta_{n}\left(\rho_{n} \star u^{\prime}\right). $$ Therefore $$ \begin{aligned} \left\|u_{n}^{\prime}-u^{\prime}\right\|_{p} & \leq\left\|\zeta_{n}^{\prime}\left(\rho_{n} \star u\right)\right\|_{p}+\left\|\zeta_{n}\left(\rho_{n} \star u^{\prime}\right)-u^{\prime}\right\|_{p} \\ & \leq \frac{C}{n}\|u\|_{p}+\left\|\rho_{n} \star u^{\prime}-u^{\prime}\right\|_{p}+\left\|\zeta_{n} u^{\prime}-u^{\prime}\right\|_{p} \rightarrow 0, \end{aligned} $$ where $C=\left\|\zeta^{\prime}\right\|_\infty$.

The next result is an important prototype of a Sobolev inequality (also called a Sobolev embedding $)$

• Theorem 8.8. There exists a constant $C$ (depending only on $|I| \leq \infty$ ) such that $$ \|u\|_{L^{\infty}(I)} \leq C\|u\|_{W^{1, p}(I)} \quad \forall u \in W^{1, p}(I), \quad \forall 1 \leq p \leq \infty. \tag{5} $$

Answer 1

You observe correctly that $\|\zeta_n\|_{L^p} \to \infty$ but this is not used. The inequality $$\|u_n - u\|_p≤ \|\rho_n \star u - u\|_{p}+\|\zeta_n u - u\|_p$$ follows by triangle inequality and the point wise bound $\|\zeta_n\|_\infty≤1$ independently of $n$. The first term is null due to properties of appropriately chosen mollifiers, and the second term is null because $u\in L^p$ and the hopefully clear picture that $(1-\zeta_n) \approx \mathbb 1_{|x|>n}$: $$\| \zeta_n u - u\|_p^p ≤ \int_{ |x|>n}|u|^p \, dx \to 0 $$ This is like the fact that the tail sum of a convergent series is null; in fact, the following decomposition provides exactly such a proof: $$\|u\|_{L^p}^p = \sum_{k=0}^\infty\int_{k≤|x|<k+1}|u|^p \, dx $$