Let \(\Omega\) be a bounded domain in \(\mathbb{R}^d\). \(H^{-1}(\Omega)\) is defined as the dual space of \(H_0^1(\Omega)\), which is the space of all bounded linear functionals on \(H_0^1(\Omega)\). Then the \(L_2(\Omega)\) space is a subspace of \(H^{-1}(\Omega)\). This can be verified as below.
Let \(f \in L_2(\Omega)\). Define a linear functional using the inner product on \(L_2(\Omega)\) as \(f(v) = (f, v)\) for all \(v \in L_2(\Omega)\). If \(v \in H_0^1(\Omega)\), \(v\) also belongs to \(L_2(\Omega)\). Therefore, the linear functional \(f\) can also be applied to \(v\). Because \[\lVert v\rVert_1^2 = (\lVert v\rVert_{L_2}^2 + \lVert\nabla v\rVert_{L_2}^2) \geq \lVert v\rVert_{L_2}^2,\] \[\lVert f\rVert_{-1} = \sup_{v \in H_0^1(\Omega), v \neq 0} \frac{\lvert f(v)\rvert}{\lVert v\rVert_1} \leq \sup_{v \in H_0^1(\Omega), v \neq 0} \frac{\lvert f(v)\rvert}{\lVert v\rVert_{L_2}} \leq \sup_{v \in L_2(\Omega), v \neq 0} \frac{\lvert f(v)\rvert}{\lVert v\rVert_{L_2}} = \lVert f\rVert_{L_2}.\] Therefore, if \(f \in L_2(\Omega)\), its \(H^{-1}\) norm is also finite, hence \(f \in H^{-1}(\Omega)\) and \(L_2(\Omega) \subset H^{-1}(\Omega)\).
To show \(H^{-1}(\Omega) \subset L_2(\Omega)\), an example function can be provided, which belongs to \(H^{-1}(\Omega)\) but not to \(L_2(\Omega)\).
Let \(\Omega = (0,1)\) and \(f(x) = \frac{1}{x}\). \(f(x)\) has a singularity at \(x\) and the integral \(\int_{\Omega} f^2 \,{\rm d}x = \int_0^1 \frac{1}{x^2} \,{\rm d}x\) is divergent. Therefore, \(f \notin L_2(\Omega)\).
Next, define \(f(v) = (f, v)\) for any \(v \in H_0^1(\Omega)\), we have \[\begin{aligned} (f, v) &= \int_0^1 \frac{1}{x} v(x) \,{\rm d}x = \int_0^1 v(x) \,{\rm d}(\log x) \\ &= v(x)\log x \big\vert_0^1 - \int_0^1 \log x \frac{{\rm d}v}{{\rm d} x} \,{\rm d}x. \end{aligned}\] Because \(v \in H_0^1(\Omega)\), the boundary term in the above integral disappears and \[(f, v) = -\int_0^1 \log x \frac{{\rm d}v}{{\rm d}x} \,{\rm d}x.\]
Then we want to apply Cauchy-Schwartz inequality to this formula. But before that, we need to verify that both \(\log x\) and \(\frac{{\rm d}v}{{\rm d}x}\) belong to \(L_2(\Omega)\). Because \(v \in H_0^1(\Omega)\), the latter belongs to \(L_2(\Omega)\) for sure. For the former, we need to show \(\int_0^1 (\log x)^2 \,{\rm d}x\) is finite.
We notice that \[\frac{{\rm d}(x\log x - x)}{{\rm d}x} = \log x + x \frac{1}{x} - 1 = \log x\] and \[\frac{{\rm d}(x\log^2 x)}{{\rm d}x} = \log^2 x + x (2\log x) \frac{1}{x} = \log^2 x + 2\log x.\] Then we have \[\frac{{\rm d}(x\log^2 x - 2x\log x + 2x)}{{\rm d}x} = \log^2 x.\] Hence, using l’Hospital’s rule, \[\int_0^1 \log^2 x \,{\rm d}x = 2\] and \(\log x \in L_2(\Omega)\). Now we apply Cauchy-Schwartz inequality, \[\lvert(f, v)\rvert = \bigg\lvert\int_0^1 \log x \frac{{\rm d}v}{{\rm d}x} \,{\rm d} x\bigg\rvert \leq \lVert\log x\rVert_{L_2} \bigg\lVert\frac{{\rm d}v}{{\rm d}x}\bigg\rVert_{L_2} = \sqrt{2} \lvert v\rvert_1.\] Therefore, \(\sup \frac{\lvert f(v)\rvert}{\lvert v\rvert_1} \leq \sqrt{2}\) and \(f \in H^{-1}(\Omega)\). \(L_2(\Omega)\) is a proper subspace of \(H^{-1}(\Omega)\).
[1] Larsson, Stig, and Vidar Thomee. 2003. Partial Differential Equations with Numerical Methods. Texts in Applied Mathematics. Berlin Heidelberg: Springer-Verlag: Problem A.10 and A.11 on p241.
$L_2(\Omega)$ is a subspace of $H^{-1}(\Omega)$
原文：https://www.cnblogs.com/peabody/p/14033793.html