We have the triple inclusions $H^1_0(\Omega)\subset L^2(\Omega)\subset H^{-1}(\Omega)$, where the second inclusion is not literal but in the sense of distributions. Related to this answer, why is the corresponding statement $\mathbb{R}\subset\mathbb{R^2}\subset\mathbb{R}$ obviously false?
Does not the second statement make sense when the second inclusion is taken in the sense of functionals?
If the second statement also make sense, does it have any uses just as the first triple inclusion does in pdes?
You can formalize this:
Take two Hilbert spaces $H\subset V$ such that
The inclusion is dense (i.e. the image of $H$ in $V$ is dense in the topology of $V$) and,
The inclusion is continuous (i.e. $H$ has a stronger norm than $V$).
Then, if we identify $V$ with its dual (let's assume the spaces are real, though you can do this in general), we have that $V\subset H^*$ in the following sense:
Every $v\in V$, defines a bounded linear functional in $H$ via $L_v(h)=(h, v)_V$; to see this notice that by Cauchy-Schwarz we have $$|L_v(h)|\leq \| v\|_V\| h\|_V \leq \| v\|_V\| h\|_H,$$ where the last inequality is the continuity of the inclusion $H\subset V$. Moreover if $v_i$ are two elements of $V$ that are mapped to the same linear functional $L_v$, then by density of $H$ in $V$ we get that $v_1=v_2$ (since equality of the associated functionals implies that $(h, v_1)_V=(h, v_2)_V$ for every $h\in H$).
Therefore we can view our Hilbert spaces as satisfying $$ H\subset V\subset H^*. $$ By the way this is usually called a Hilbert triple. I'm sure if you search for it you can find more detailed info on this topic.
As for why this fails in the case $\mathbb{R} \subset \mathbb{R}^2$, notice that we don't have a dense inclusion, and so we can't conclude that distinct elements in $\mathbb{R}^2$ are sent to distinct elements of $\mathbb{R}^*$. In fact if we have a Hilbert triple, then all spaces are infinite dimensional or $V=H$.