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If $f\in L^2(\Omega)$, where $\Omega$ is a domain in $\mathbb{R}^n$, why is it that the distributional of $f$, say with respect to $x_1$, is in $H^{-1}(\Omega)$, the dual space of $H^1_{0}(\Omega)$?

I tried to make sense of it by writing out the definition of distributional derivative, but I could not see why $\partial x_1 f $ is in $H^{-1}$

mononono
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1 Answers1

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For $\varphi \in C^{\infty}_c$ one defines $\langle\partial_{x_1}f,\varphi\rangle = -\int_{\Omega}f\partial_{x_1}\varphi$. Extending this definition to $H^1_0(\Omega)$ we obtain a linear functional on $H^1_0(\Omega)$. Continuity, i.e. boundedness, follows from Holder's inequality (together with Poincare's inequality, depending on the norm you define on $H^1_0$): $$\Big|\int f\partial_{x_1}\varphi\Big|\le \|f\|_{L^2}\|\nabla\varphi\|_{L^2} \Big(\le C\|f\|_{L^2}\|\varphi\|_{H^1_0}\Big).$$

Giovanni
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  • When I extend the definition to a function $u \in H^1_0(\Omega)$, I suppose that I need to use the density of $C^{\infty}c$ in $H^1_0$ by considering $u_n \to u$ in $H^1$, $u_n$ smooth. But, how do I know that $\lim<\partial{x_1}f,u_n>$ exists? – mononono Feb 01 '16 at 23:31
  • that's part of the definition of $H_0^1(\Omega)$ that $||\partial_{x_1} \varphi||_{L^2}$ is finite – reuns Feb 01 '16 at 23:35
  • You can use the same definition but with $\varphi \in H^1_0$ and it is well defined thanks to the integrability condition built in $H^1_0$ – Giovanni Feb 01 '16 at 23:38
  • $\varphi \in H_0^1(\Omega)$ is precisely the set for which $\int_{\Omega} |f(x) \varphi(x)|^2 + |g(x) \varphi'(x)|^2 dx$ is finite for any $f,g \in L_2$ which implies $\int_{\Omega} f(x) \varphi'(x) dx$ exists – reuns Feb 01 '16 at 23:43