Inclusion of dual space: $X\subset Y \implies Y^*\subset X^*$

Question

I'm confused at dual space of Hilbert space. Perhaps it is easy, but I don't know, please help me.

Let $X,$ $Y$ be Hilbert spaces and $X^\ast$, $Y^\ast$ are the dual spaces of $X$ and $Y$ respectively. Suppose that $X\subset Y$. Then, $Y^\ast\subset X^\ast$.

This is correct according to books about functional analysis. But, I have no idea to confirm this inclusion. I think $X^\ast\subset Y^\ast$, because the domain of $Y^\ast$ is bigger than the one of $X^\ast$. Why this idea is incorrect?

Answer 1

In general, strictly speaking, unless $X = Y$, the functionals in $X^*$ and $Y^*$ are defined over different domains, and hence $X^* \cap Y^* = \emptyset$. That is, it is not true that $X^* \subseteq Y^*$ or $Y^* \subseteq X^*$. This applies not just to Hilbert spaces, but to any normed (or indeed topological) vector spaces.

This is a particularly glib interpretation of the question, but it's difficult to answer more rigorously without knowing exactly what liberties we can take when talking about the $\subseteq$ relation.

Answer 2

Considering Hilbert (sub)spaces $X\subset Y$, we know that every functional in $X^*$ can be mapped into $Y^*$ by restriction of domain. Since $X\neq Y$, it can be shown using orthogonal compliments that this restriction is surjective, but not injective. This is one sense in which $Y^*$ is "smaller than" $X^*$, though the comparison $Y^*\subset X^*$ is not meaningful without clarification.

Answer 3

I guess that in this context $X \subset Y$ means that there is an injective, linear and continuous map $i \colon X \hookrightarrow Y$. $i$ induces a map $i^* \colon Y^* \to X^*, \varphi \mapsto \varphi \circ i$ (“restricting to $X$”). This map is linear and continuous but in general not injective. However, $i^*$is injective (i.e. $Y^* \subset X^*$) under the additional assumption that the image of $i$ is dense in $Y$.

Together with the Riesz isomorphism $Y \cong Y^*$ we can interpret this result as $X \subset Y \subset X^*$. This is called a Gelfand and triple and is important e.g. in the theory’s of PDEs.

Answer 4

I suppose that the inner product on $X$ is the restriction of the inner product $(\cdot, \cdot)$ on $Y$.

Now let $f \in Y^\ast$. Then there is $y \in Y$ such that $f(z)=(z,y)$ for all $z \in Y.$

Hence $f(x)=(x,y)$ for all $x \in X.$

Conclusion: for the restriction $f_{|X}$ we have $f_{|X} \in X^\ast.$