Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I don't see how to solve this problem which I think should be easy:
Let Y be a reflexive space. Assume $Y$ is continuously embedded in a Hilbert space $H$ and $Y$ is dense in $H$. Show that $H^*$ is dense in $Y^*$. I already know that $H^*$ is continuously embedded in $Y^*$ so you may use that if it helps.
Any help would be appreciated, thanks.
Let $T:Y\to H$ be the embedding operator and $T^*:H^*\to Y^*$ its adjoint. Suppose the range of $T^*$ is contained in a proper closed subspace of $Y^*$. There is a bounded linear functional on $Y^*$ that vanishes on that subspace. By reflexivity, this functional is evaluation on some nonzero element $y\in Y$. Let's spell out what this means: for every $\varphi\in H^*$ the composition $\varphi\circ T$ vanishes on $y$. This is only possible if $Ty=0$, but $T$ is injective. $\quad \Box$
Note that the density of $Y$ in $H$ is not needed for this part of the argument; you needed it for $T^*$ to be injective.
