# Why $H^1_0(U) \subset L^2(U) \subset H^{-1}(U)$?

I'm reading by myself the PDE's book by Lawrence Evans and he denoted the dual of $$H^1_0(U)$$, which is the closure of $$\mathcal{C}^{\infty}_c(U)$$ in $$W^{1,2}(U)$$ ($$H^1_0(U) = W^{1,2}_0(U)$$ ), by $$H^{-1}(U)$$. In what follows, Evans states

$$\textit{Note very carefully that we do not identify the space}$$ $$H^1_0(U)$$ $$\textit{with its dual}$$. Instead, as we will see in a moment, we have

$$H^1_0(U) \subset L^2(U) \subset H^{-1}(U). (*)$$

I can see that $$H^1_0(U) \subset L^2(U)$$ since $$H^1_0(U) = \overline{\mathcal{C}^{\infty}_c(U)} \subset W^{1,2}(U) \subset L^2(U)$$, but I can't see why $$L^2(U) \subset H^{-1}(U)$$ since $$H^{-1}(U)$$ is the space of bounded linear functions and I don't know if all $$f \in L^2(U)$$ is a linear function. I understand $$(*)$$ literally as a space contained in another space, but the author of this OP led me to think that $$(*)$$ maybe can be true doing identifications, as the author of the topic did, so my question is if $$(*)$$ is valid doing identifications or not. If $$(*)$$ is valid by continence of spaces, then why $$L^2(U) \subset H^{-1}(U)$$?

You should make some kind of identification, because the objects you are taking into account are logically different: $$L^2$$ is the space of equivalence classes of certain functions from $$\mathbb{R}$$ to $$\mathbb{R}$$, while $$H^{-1}$$ is a space of functions from $$H^1_0$$ to $$\mathbb{R}$$. The point is that there is a natural identification of a function $$f\in L^2$$ with a dual function $$H^1_0 \to \mathbb{R},\ g \mapsto \int_\Omega fg = (f,g)_{L^2}$$ This identification is natural as it is induced by the scalar product on $$L^2$$, which is "part of" the scalar product on $$H^1$$. Notice that the idenfication we are considering is $$f \eqsim (f,\cdot)$$, which is indeed in $$H^{-1}$$ since $$(f,g) = \int_\Omega fg \leq \|f\|_2 \|g\|_2 \leq \|f\|_2 \|g\|_{H^1_0}.$$