Dense inclusions of Banach spaces and their duals This seems like a really simple question, but I'm struggling with it.  Let $X$ be a separable Banach space, $H$ be a separable Hilbert space, and suppose $i : H \hookrightarrow X$ is a dense, continuous embedding of $H$ into $X$.  (This is the abstract Wiener space construction due to Gross, hence the [pr.probability] tag)  If we associate $H$ with its dual $H^{\star}$, we have the inclusions $$X^{\star} \hookrightarrow H^{\star} \cong H \hookrightarrow X.$$
My question:  Is $i^{\star} : X^{\star} \hookrightarrow H^{\star}$ a dense injection?
 A: Yes, if you mean that $i$ is one to one, for an operator $T:X\to Y$ is one to one if and only if $T$* has weak* dense range, which means $T$* has dense range when $X$ is reflexive.
A: Just to flesh out Bill's answer and comments thereon, we have the following facts.  Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator.


*

*If $T$ has dense range then $T^*$ is injective.


Since this is a standard homework problem I'll just give a hint. Suppose $f \in Y^*$ with $T^*f=0$.  This means that $f(Tx)=0$ for every $x$, i.e. $f$ vanishes on the range of $T$...


*Suppose further that $X$ is reflexive.  If $T$ is injective then $T^*$ has dense range.


Hint: By Hahn-Banach, to show that $T^*$ has dense range, it suffices to show that if $u \in X^{**}$ vanishes on the range of $T^*$, then $u=0$.  And by reflexivity, $u \in X^{**}$ is represented by some $x \in X$...
Note that the proofs I have in mind don't need to discuss the weak-* topology (at least not explicitly).
We cannot drop reflexivity in 2.  Consider the inclusion of $\ell^1$ into $\ell^2$.  It is injective, but its adjoint is the inclusion of $\ell^2$ into $\ell^\infty$, whose range is not dense.
(Fun fact: you can't prove 2 without Hahn-Banach or some other consequence of the axiom of choice.  If you work in ZF + dependent choice (DC), it's consistent that $\ell^1$ is reflexive, but we still have the injective map from $\ell^1$ to $\ell^2$ whose adjoint doesn't have dense range.  So under those axioms it's consistent that 2 is false.)
