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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?

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up vote 5 down vote accepted

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.

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Bill, thanks so much for the response. That answers my question; could you please give me an argument or a citation so I could look it up? – Tom LaGatta Feb 11 '10 at 2:18
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A linear subspace of a Hilbert space is dense iff its orthogonal is {0}... – Ady Feb 11 '10 at 2:28
    
Simple, as I thought. Thanks, Ady! – Tom LaGatta Feb 11 '10 at 2:39
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Just to add that Bill's answer for general Banach spaces can probably be found in several functional analysis textbooks, but can definitely be found in Chapter 4 of Rudin's FA. (Thm 4.14 in my copy, 2nd ed) – Yemon Choi Feb 11 '10 at 3:49
    
It is also in many Real Analysis textbooks. See, e.g., problem 22 after section 5.2 in Folland's Real Analysis, 2nd ed. – Bill Johnson Feb 11 '10 at 17:19

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.

  1. 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 $T$...

  1. 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.)

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