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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 $H^*$, we have the inclusions $$X^* \hookrightarrow H^* \cong H \hookrightarrow X.$$

My question: Is $i^* : X^* \hookrightarrow H^*$ a dense injection?

edited Feb 11 2010 at 2:09

asked Feb 11 2010 at 1:55

Tom LaGatta
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Bill Johnson · Accepted Answer · 2010-02-11 02:11:00Z UTC

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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.

answered Feb 11 2010 at 2:11

Bill Johnson
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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 2010 at 2:18

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A linear subspace of a Hilbert space is dense iff its orthogonal is {0}... – Ady Feb 11 2010 at 2:28

Simple, as I thought. Thanks, Ady! – Tom LaGatta Feb 11 2010 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 2010 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 2010 at 17:19

Dense inclusions of Banach spaces and their duals

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