On convergence of particular functions in Sobolev space

Question

I'm having trouble showing the following fact:

Suppose $1\le p<\infty$, $\ \varphi\in C^\infty(0,\infty)$ such that $$\varphi(s)= \begin{cases} 0, & s\leq 1/2\\ y\in[0,1],& s\in(1/2,1) \\ 1,& s\geq 1 \end{cases}$$ and $v\in C^1(\mathbb R \times (0,\infty))\, \cap W^{1,p}(\mathbb R\times (0,\infty))$ such that $v(x_1, 0) = 0$ for all $x_1\in\mathbb{R}$.

Then $$ \lim\limits_{\varepsilon\to +0}\left\| v(x_1, x_2) \varphi(x_2/\varepsilon) - v(x_1,x_2)\right\|_{W^{1,p}\left(\mathbb R\times (0,\infty)\right)} = 0 $$

Would you please help me?

Answer 1

Let's formulate a lemma:

Suppose $f\in L^p(\mathbb{R}\times(0,+\infty))$ and $\varphi\in C^\infty(0,+\infty)$ as in the question. Then $$\lim\limits_{\varepsilon\to 0}\|f(x_1,x_2)\varphi(x_2/\varepsilon)-f(x_1,x_2)\|_{L^p}=0$$

Proof of the lemma. Note that $$ \int\limits_{\mathbb{R}\times(0,+\infty)}|f(x_1,x_2)\varphi(x_2/\varepsilon)-f(x_1,x_2)|^pd\mu = $$ $$ =\int\limits_{\mathbb{R}\times(0,\varepsilon)}|f(x_1,x_2)\varphi(x_2/\varepsilon)-f(x_1,x_2)|^pd\mu\ \ \leq\int\limits_{\mathbb{R}\times(0,\varepsilon)}|f(x_1,x_2)|^pd\mu $$ And the latter goes to $0$ as $\varepsilon\to +0$, since $\int_{\mathbb{R}\times(0,\varepsilon)}|f(x_1,x_2)|^pd\mu \leq\|f\|^p_{L^2(\mathbb{R}\times(0,+\infty))}<\infty$.

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Now using this lemma we have first $$\lim\limits_{\varepsilon\to 0}\|v(x_1,x_2)\varphi(x_2/\varepsilon)-v(x_1,x_2)\|_{L^p}=0,$$ second $$\lim\limits_{\varepsilon\to 0}\left\|\frac{\partial v}{\partial x_1}(x_1,x_2)\varphi(x_2/\varepsilon)-\frac{\partial v}{\partial x_1}(x_1,x_2)\right\|_{L^p}=0,$$ and third $$\lim\limits_{\varepsilon\to 0}\left\|\frac{\partial v}{\partial x_2}(x_1,x_2)\varphi(x_2/\varepsilon)+\frac{1}{\varepsilon}v(x_1,x_2)\varphi'(x_2/\varepsilon)-\frac{\partial v}{\partial x_2}(x_1,x_2)\right\|_{L^p}\leq$$ $$ \leq \underbrace{\lim\limits_{\varepsilon\to 0}\left\|\frac{\partial v}{\partial x_2}(x_1,x_2)\varphi(x_2/\varepsilon)-\frac{\partial v}{\partial x_2}(x_1,x_2)\right\|_{L^p}}_{=0\ \textrm{by lemma}} + \lim\limits_{\varepsilon\to 0}\frac{1}{\varepsilon}\left\|v(x_1,x_2)\varphi'(x_2/\varepsilon)\right\|_{L^p}= $$ $$ = \lim\limits_{\varepsilon\to 0}\frac{1}{\varepsilon}\left\|v(x_1,x_2)\varphi'(x_2/\varepsilon)\right\|_{L^p} \leq \lim\limits_{\varepsilon\to 0}\left(\frac{1}{\varepsilon}\left\|v(x_1,x_2)\right\|_{L^p}\cdot\sup\limits_{(0,+\infty)}\varphi'\right)=0. $$ All these three gives us $$ \lim\limits_{\varepsilon\to 0}\left\| v(x_1, x_2) \varphi(x_2/\varepsilon) - v(x_1,x_2)\right\|_{W^{1,p}\left(\mathbb R\times (0,\infty)\right)} = 0 $$