$L_2(\Omega)$ is a subspace of $H^{-1}(\Omega)$
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 \[ \norm{v}_1^2 = (\norm{v}_{L_2}^2 + \norm{\nabla v}_{L_2}^2) \geq \norm{v}_{L_2}^2, \] \[ \norm{f}_{-1} = \sup _{v \in H_0^1(\Omega ), v \neq 0} \frac{\abs{f(v)}}{\norm{v}_1} \leq \sup _{v \in H_0^1(\Omega ), v \neq 0} \frac{\abs{f(v)}}{\norm{v}_{L_2}} \leq \sup _{v \in L_2(\Omega ), v \neq 0} \frac{\abs{f(v)}}{\norm{v}_{L_2}} = \norm{f}_{L_2}. \] 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 \(\int _{\Omega } f^2 \intd x = \int _0^1 \frac{1}{x^2} \intd x\) 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{equation*} \begin{aligned} (f, v) &= \int _0^1 \frac{1}{x} v(x) \intd x = \int _0^1 v(x) \intd (\log x) \\ &= v(x)\log x \big \vert _0^1 - \int _0^1 \log x \frac{\diff v}{\diff x} \intd x. \end{aligned} \end{equation*} Because \(v \in H_0^1(\Omega )\), the boundary term in the above integral disappears and \[ (f, v) = -\int _0^1 \log x \frac{\diff v}{\diff x} \intd x. \]
Then we want to apply Cauchy-Schwartz inequality to this formula. But before that, we need to verify that both \(\log x\) and \(\frac{\diff v}{\diff x}\) 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 \(\int _0^1 (\log x)^2 \intd x\) is finite.
We notice that \[ \frac{\diff (x\log x - x)}{\diff x} = \log x + x \frac{1}{x} - 1 = \log x \] and \[ \frac{\diff (x\log ^2 x)}{\diff x} = \log ^2 x + x (2\log x) \frac{1}{x} = \log ^2 x + 2\log x. \] Then we have \[ \frac{\diff (x\log ^2 x - 2x\log x + 2x)}{\diff x} = \log ^2 x. \] Hence, using l’Hospital’s rule, \[ \int _0^1 \log ^2 x \intd x = 2 \] and \(\log x \in L_2(\Omega )\). Now we apply Cauchy-Schwartz inequality, \[ \abs{(f, v)} = \bigabs{\int _0^1 \log x \frac{\diff v}{\diff x} \intd x} \leq \norm{\log x}_{L_2} \bignorm{\frac{\diff v}{\diff x}}_{L_2} = \sqrt{2} \abs{v}_1. \] Therefore, \(\sup \frac{\abs{f(v)}}{\abs{v}_1} \leq \sqrt{2}\) and \(f \in H^{-1}(\Omega )\). \(L_2(\Omega )\) is a proper subspace of \(H^{-1}(\Omega )\).
