Let $\Omega\subset\mathbb{R}^N$ be a bounded open set. I was wondering if $C_c^\infty(\Omega)$ is dense in $W^{1,p}(\Omega)\cap C_0(\Omega)$ with $1\leq p<\infty$.

Here $C_0(\Omega)$ is the space of functions of $C(\overline{\Omega})$ which are zero on $\partial\Omega$, and the space $W^{1,p}(\Omega)\cap C_0(\Omega)$ is endowed with the norm $\|\cdot\|_{W^{1,p}}+\|\cdot\|_{L^\infty}$.

Thanks in advance.


1 Answer 1


Take a function $f$ and convolve with a Mollifier $\phi_{\epsilon}$ and convert it into compactly supported smooth function $(1_{B_r(0)} \times f) * \phi_{\epsilon}$ where $1_{B_r(0)}$ is an indicator function of compact set $Closure(B_r(0))$. It is known that $$ \lim_{\epsilon \rightarrow 0} \lim_{r \rightarrow \infty} (1_{B_r(0)} \times f) * \phi_{\epsilon} \rightarrow f.$$ Since $\Omega$ is a bounded set, we dont have to worry about compactness since $f|_{\partial \Omega} = 0$. We now prove that closure of $C^{\infty}(\Omega + B_{\epsilon}(0))$ (when functions are restricted to $\Omega$) contains $W^{1,p}(\Omega) \cap C_0(\Omega)$. Hence enough to prove: $$f * \phi_{\epsilon} \rightarrow f$$ Let $f \in C_0(\Omega)$, $$f_{\epsilon}(x)-f(x) = \int (f(x-y)-f(x)) \phi_{\epsilon}(y) dy $$ $$|f_{\epsilon}(x)-f(x)| \leq \int |f(x-y)-f(x)| \phi_{\epsilon}(y) dy $$ Since $f$ is uniformly continuous as $\Omega$ is a bounded set and $f|_{\partial \Omega} = 0$ and since $\phi_{\epsilon}$ is supported in the set $B_{\epsilon}(0)$, we have by choosing $\epsilon$ sufficiently small: $$|f_{\epsilon}(x)-f(x)| \leq \epsilon_f \int \phi_{\epsilon}(y) dy $$ $$|f_{\epsilon}(x)-f(x)| \leq \epsilon_f$$ Hence $f_{\epsilon} \rightarrow f$ uniformly.

By uniform convergence, we also have that, $$\int |f_{\epsilon}(x)-f(x)|^p dx \leq \epsilon_f^p \times \mu(\Omega)$$ Hence $f_{\epsilon} \rightarrow f$ in $L_p$.

Now observe that: $$f'_{\epsilon}(x) = \int f'(x-y) \phi_{\epsilon}(y) dy $$ and we repeat the same proof as above to conclude that $f_{\epsilon} \rightarrow f$ in $||.||_{W^{1,p}} + ||.||_{L_{\infty}}$. To repeat the proof, we might need the condition $f'$ is continuous and $f'|_{\partial \Omega} = 0$ OR $f'$ is uniformly continuous. Try this out and let me know.

Hence any $f_{\epsilon} \rightarrow f$ for any $f \in W^{1,p}(\Omega) \cap C_0(\Omega)$

You can also find some materials in: http://wwwarchive.math.psu.edu/anovikov/acm105/mollifiers.pdf

EDIT: The idea is to work with $\Omega'$ where $\Omega'$ is slightly larger than $\Omega$ such that $Closure(\Omega) \subset M \subset \Omega'$ where $M$ is a compact set. Extend the $L_p$ function $f'$ to $\Omega'$. we need to set $f'(x) = 0$, $\forall x \in \Omega' \setminus \Omega$. Using density of $C(\Omega')$ in $L^p(\Omega')$, one can approximate $f'$ by a continuous function $g$ on $\Omega'$ and approximate the continuous function $g$ by smooth function $g_{\epsilon}$ on $Closure(\Omega)$ by using uniform continuity of $g$ on $M$ and use the inequality $|f'_{\epsilon}-g_{\epsilon}|_{L_p(\Omega)} \leq |f'-g|_{L_p(M)}$ and show that $f'_{\epsilon} \rightarrow f'$ on $\Omega$ by triangle inequality. Thus completing the proof by observing $|f'-g|_{L_p(\Omega)} \leq |f'-g|_{L_p(M)} \leq |f'-g|_{L_p(\Omega')}$.
The following link has some of the above details. See Theorem 1.28,Page 18,19, in https://www.math.ucdavis.edu/~hunter/m218a_09/Lp_and_Sobolev_notes.pdf

EDIT: In the above we have proved that closure of $C^{\infty}( \Omega + B_{\epsilon}(0))$ (when functions are restricted to $\Omega$) contains $W^{1,p} \cap C_0(\Omega)$. Please see below for the argument that $C_{c}^{\infty}(\Omega)$ (smooth function on compact set) is dense.

Its enough to prove $(f 1_K) * \phi_{\epsilon} \rightarrow f * \phi_{\epsilon}$ i.e., $(f 1_{\Omega \setminus K}) * \phi_{\epsilon} \rightarrow 0$. Now $$|(f 1_{\Omega \setminus K}) * \phi_{\epsilon}| = |\int (f 1_{\Omega \setminus K})(x-y) * \phi_{\epsilon}(y) dy| \leq \int |(f 1_{\Omega \setminus K})(x-y)| * \phi_{\epsilon}(y) dy.$$ Since $f|_{\partial \Omega} = 0$ and $\Omega$ is a bounded set, by uniform continuity of $f$, there exists $K = K_{\gamma, \epsilon} \subset \Omega$, $K_{\gamma, \epsilon}$ compact such that $|f(x)| \leq \gamma \epsilon^{n+1}$, $\forall x \in \Omega \setminus K_{\gamma, \epsilon}$. Note that $|\phi_{\epsilon}(y)| \leq 1/\epsilon^n$ (see pdf in the link above). Using this, we get: $$|(f 1_{\Omega \setminus K}) * \phi_{\epsilon} (x)|\leq \int |(f 1_{\Omega \setminus K})(x-y)| * \phi_{\epsilon}(y) dy \leq \gamma \epsilon \mu(\Omega).$$ To prove derivative, converges use exactly same argument with $\phi'_{\epsilon}$ in place of $\phi_{\epsilon}$ and use $|\phi'_{\epsilon}(y)| \leq 1/\epsilon^{n+1}$.

EDIT: As @rubik pointed out in the comments the proof is correct/complete when we have $$\mu(\Omega \setminus K_{\frac{3}{2} \times d_{\gamma}})^{1/p} \leq constant \times d_{\gamma}.$$ as $d_{\gamma} \rightarrow 0$, where we can define: $K_{d} = \{x: x \in \Omega, dist(x,\partial \Omega) \geq d\}$ and set $K_{\gamma,\epsilon} = K_{d_{\gamma}}$ for $d_{\gamma}$ which depends on continuity of $f$ at the boundary. This is true atleast for open balls/rectangles of any radius $r$ in any $N$ dimensional space when $p=1$.

  • $\begingroup$ Thanks for your ideas and materials. It is true that the function $f_\epsilon=f*\phi_\epsilon$ with $\phi_\epsilon$ the standard mollifier satisfies $f_\epsilon\to f$ in $W^{1,p}(\Omega)\cap C_0(\Omega)$. However, $f_\epsilon\in C^\infty(\mathbb{R}^N)$, not in $C_c^\infty(\Omega)$. In this way, it seems that one needs to think about how to approximate $C^\infty$ by $C_c^\infty$ in the $W^{1.p}$ sense (the problem mainly lies in the approximation of the gradient). $\endgroup$
    – rubik
    Nov 28, 2022 at 11:05
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    $\begingroup$ This seems true for open balls/rectangles of radius< 1, for any $N$, for $p=1$ as $d \rightarrow 0$. So i think what you want is true for $p=1$ for these specific domains (Balls/rectangles of radius <1). $\endgroup$
    – Balaji sb
    Dec 1, 2022 at 4:40
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    $\begingroup$ Thanks for your update. Indeed, it seems to be "yes" for balls or rectangles. Focusing on the ratio $|\Omega-K|^{1/p}/d$ appeared in our derivation, it seems to be possibly bounded only if $p=1$. Actually, take $K=\{x\in\Omega:d(x)>\epsilon\}$ where $d(x)=dist(x,\partial\Omega)$. By the coarea formula, $|\Omega-K|=\int_0^\epsilon\mathcal{H}^{N-1}(d^{-1}(t))$. By the isoperimetric ineq, $\mathcal{H}^{N-1}(d^{-1}(t))\ge C(n)|\{x\in\Omega:d(x)>t\}|^{(N-1)/N}\ge C(|K|,n)$ for a.e. t; So $|\Omega-K|^{1/p}/\epsilon\ge C(n,|K|)\epsilon^{1/p-1}$, which is possibly bounded only if $p=1$. $\endgroup$
    – rubik
    Dec 4, 2022 at 6:37
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    $\begingroup$ By the way, I'd like to mention that the mollification process $f*\phi_\epsilon\to f$ in $W^{1,p}$ seems not necessarily valid for any $\Omega$. Since during the process, we regarded $f$ as 0 outside $\Omega$. This extension can't guarantee $Df$ is still in $L^p$. So, generally, that approximation of the gradient works locally. You can check a relevant book to see how to construct a global approximation (partition of unity would be used, see the result that "$C^\infty$ is dense in $W^{1,p}$"). $\endgroup$
    – rubik
    Dec 4, 2022 at 9:04
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    $\begingroup$ Anyway, the above process works if we know the trivial extension is valid for $Df$, like when $f\in W_0^{1,p}$. This can be guaranteed by assuming $\Omega$ to be Lipschitz, for which the trace characterization tells that $W^{1,p}\cap C_0\subset W_0^{1,p}$. In addition, if $p>N$ then it holds the Sobolev embedding $W_0^{1,p}\hookrightarrow C^{0,\alpha}(\overline{\Omega})$. Clearly, this fact could affirmatively answer the question. As for arbitrary $\Omega$, the corresponding conclusions remain to explore. Let's expect to see that soon. $\endgroup$
    – rubik
    Dec 4, 2022 at 9:26

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