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Let $u>0$ be in $H^{1}(\mathbb{R})=W^{1,2}(\mathbb{R})$, we know that the set of $C^{\infty}$ functions with compact support are dense in the Sobolev space $H^{1}(\mathbb{R})$. Hence, we have a sequence $u_n$ (convolution +cut-off) which converges toward to $u$ using the $H^1(\mathbb{R})$ norm.

Can I impose the condition $u_n(x)\leq u(x)$ a.e?

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    $\begingroup$ If you insist on constructing it from convolution against a mollifier: no. But if you just want a sequence of smooth approximations: yes. $\endgroup$
    – Willie Wong
    Commented Jul 26, 2016 at 15:00
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    $\begingroup$ Rough sketch: since $u > 0$, you can find a sequence $R_n \nearrow \infty$ and $\epsilon_n \searrow 0$ such that the functions $$ v_n = \chi_{[-R_n, R_n]}\cdot (u - \epsilon_n) $$ is non-negative and converges to $u$ in $H^1$. For each $v_n$ you can mollify. Then you can diagonalize and use the fact that $u\in H^1 \implies u$ is uniformly continuous to ensure the smoothed versions of $v_n$ remains below $u$. $\endgroup$
    – Willie Wong
    Commented Jul 26, 2016 at 15:15
  • $\begingroup$ Thank for your answer, it is not necessary to have a convolution against mollifier. I just want a sequence of $C^{\infty}(\mathbb{R})$ with compact support and the condition $|u_n(x)|\leq u(x)$ a.e. $\endgroup$
    – papnass
    Commented Jul 26, 2016 at 18:17
  • $\begingroup$ Can you please explain more how to transform the functions $v_n$ into a functions in $C^{\infty}_{0}(\mathbb{R})$ which converge toward to $u$ using the $H^{1}$ norm and maintain the condition $|v_n(x)|\leq u(x)$ a.e. $\endgroup$
    – papnass
    Commented Jul 26, 2016 at 18:32
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    $\begingroup$ Do the usual thing by convolving with a smooth bump function; call the smoothed versions $v_{n,\epsilon}$. Since $v_{n,\epsilon} \to v_n$ in $H^1$ as $\epsilon \to 0$, and $H^1(\mathbb{R})$ embeds in $L^\infty$, the $\epsilon_n$ room you left in step one is big enough. // You can be even more precise by using uniform continuity to show that as long as your smooth bump function has sufficiently small support this will work. $\endgroup$
    – Willie Wong
    Commented Jul 26, 2016 at 18:36

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As stated in some of the comments.

Let $0 \le \eta_n (x) \le 1$ be a sequence of smooth functions $\eta_n = 1$ in $[-n, n]$ and $\eta_n \in [-(n+1), n+1]$. Simply take $u_n = \eta_n \cdot u$. Basically by taking a smooth transition function in $(0,1)$.

Then $$ \| u_n' - u' \|_{L^2} \le \| \eta_n ' u \|_{L^2} + \| \eta_n u' - u' \|_{L^2} \le C \left( \int_{n <|x| < n + 1} |u|^2 \right)^{\frac 1 n} + \left( \int_{|x|>n} |u'|^2 \right)^{\frac 1 n} \to 0 $$ as $n \to \infty$. The $\| u_n - u \|_{L^2} \to 0$ I leave for you.

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