Does $\iota: V\longrightarrow H$ densely imply $\iota^\top:H'\longrightarrow V'$ dense?

Question

I was wondering if $H$ is a Hilbert space, $V$ is a Banach space with its own norm but dense in $H$ with respect to the norm of $H$, and the natural embedding $\iota:V\longrightarrow H$ is continuous, is the transpose of $\iota$, i.e.,

$$\iota^{\top}:H'\longrightarrow V'$$ also dense? So basically is $\text{Ran}(\iota^\top)\subset V'$ dense? Note that embedding only means linear, continuous and injective, it doesn't need to be a isomorphism on its range.

If it isnt the case, does it change when I consider $V$ as a Hilbert space aswell?.

Answer 1

The comments correctly gave a counterexample for the case where $V$ is just a Banach space. I think that I proved that in the case when $V$ is a reflexive Banach space the density of $\iota^\top$ is present even if $\iota$ isn't dense! (But still continuous and injective)

Proof:

Suppose $\overline{\text{Ran}(\iota^\top)}\ne V'$, then we can choose $\alpha \in \overline{\text{Ran}(\iota^\top)}^C$ for which the Hahn-Banach theorem gives us a $f \in V''$ such that:

• $f(\varphi)=0$ for $\varphi \in \overline{\text{Ran}(\iota^\top)}$
• $f(\alpha)>0$

Now for any $\varphi\in \overline{\text{Ran}(\iota^\top)}$

$$0<f(\alpha)=f(\alpha)-f(\varphi)=f(\alpha-\varphi)=(\alpha-\varphi)(u)=\alpha(u)-\varphi(u)$$

for a certain $u\in V\setminus\lbrace 0\rbrace$ (Because $V$ is reflexive).

We choose

$$\frac{\alpha(u)}{\|\iota(u)\|_H^2}\iota(u)=:\psi\in H.$$

(Note $\iota$ is continuous and injective!)

So $\iota^\top(\psi)\in \overline{\text{Ran}(\iota^\top)}$ but

$$0<\alpha(u)-\iota^\top(\psi)(u)=\alpha(u)-\iota^\top\left(\frac{\alpha(u)}{\|\iota(u)\|_H^2}\iota(u)\right)(u)=\alpha(u)-\left(\frac{\alpha(u)}{\|\iota(u)\|_H^2}\iota(u),\iota(u)\right)_H=0$$

which is a contradiction.

(Note: $\iota^\top(f)=\iota^\top((f,\cdot)_H)=(f,\iota(\cdot))_H$ via the Riesz rep theorem)