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A domain $\Omega$ is said to admit an $L_p$-Hardy inequality if there exists a finite uniform constant $C > 0$ so that the estimate

$$\int_\Omega \frac{| f(x) |^p}{d(x)^p} dx \leq \int_\Omega | \nabla f(x) |^p dx$$

holds for all functions in W^{1,p}_0(\Omega). Here d(x) means the Euclidean distance to the boundary of the domain.

Pretty much every paper on the subject begins by saying that it suffices to prove that the inequality holds for all functions in $C^\infty_0(\Omega)$ because of the density of $C^\infty_0(\Omega)$ in $W^{1,p}_0(\Omega)$. My question is why does the density of $C^\infty_0(\Omega)$ in $W^{1,p}_0(\Omega)$ imply the above result?

Davide Giraudo
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Adam Ward
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1 Answers1

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In general, if you have a topological space $X$ with a dense subset $E$, and if $\phi,\psi:X\to\mathbb R$ are continuous functions, then the validity of $\phi\le \psi$ on $E$ implies that it holds on all of $X$. Passing to the limit preserves nonstrict inequalities.

Here, the space $X$ is $W_0^{1,p}$. The righthand side, $\int |\nabla f|^p$, is continuous on $W_0^{1,p}$, by the definition of the Sobolev norm. The lefthand side looks suspicious though, with $d(x)^p$ in the denominator. Okay then, consider the inequality $$\int_\Omega \frac{| f(x) |^p}{d(x)^p+\epsilon} dx \leq C\int_\Omega | \nabla f(x) |^p dx \tag{1}$$ instead. Now both sides are continuous in the Sobolev norm. If the $L^p$-Hardy holds on $C_c^\infty$, then (1) holds on $C_c^\infty$, with the same constant $C$. Hence (1) holds on all $W_0^{1,p}$. Take the limit $\epsilon\to 0$ using the monotone convergence theorem.

user127096
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