Prove a density result in the usual Hilbert triple

Question

While studying the Hilbert triple, I've found the usual warning with identifications of dual spaces.

Let $V \subset H$, two Hilbert spaces with $V$ dense in $H$ and continuously embedded in $H$. The inner product in $H$ is denoted with $(\cdot, \cdot)_H$, the one on $V$ with $(\cdot, \cdot)_V$. By Riesz representation theorem, we can identify $H$ with its dual $H^{\star}$. I've been asked to prove that:

$H$ is dense in $V^{\star}$

Here's my attempt:

I conside a functional $T \in V^{\star}$. By Riesz, we know there's a $w \in V$ s.t. $$T(v)=(v,w)_V$$ for every $v \in V$.

Now, I pick a sequence $\{w_k \} \in V$ such that $w_k \rightarrow w$.

Then, I consider the functional $T_{w_k}:H \rightarrow \mathbb{R}$ defined as $T_{w_k}(u) = (w_k,u)_H$, which is an element of $H^{\star}$ by Schwarz inequality, as $H$ is Hilbert.

Now I consider the $$||T - T_{w_k}||_{V^{\star}}$$ and I want to show this goes to $0$. By definition of dual norm, and given a $v \in V$: $$(T-T_{w_k})(v)=(w-w_k,v)_V \rightarrow 0$$ as $w_k \rightarrow w$.

So I've found that given a functional $T \in V^{\star}$, I can find a sequence of functionals $T_k \in H^{\star}$ s.t. $$||T-T_k||_{V^{\star}} \rightarrow_k 0$$

$\square$

Is that correct?

@JeanMarie The one defined in this question (https://math.stackexchange.com/questions/251857/a-few-questions-about-the-hilbert-triple-gelfand-triple?rq=1). Basically, we have $V \subset H$ with $V$ dense in $H$ with continuous inclusion — bobinthebox, Nov 15 '21 at 13:59
you're welcome! Do you think my argument is correct?@JeanMarie — bobinthebox, Nov 15 '21 at 14:03
Shouldn't $T-T_{w_k}$ have one term which is an $H$ inner product and one term that is a $V$ inner product? — podiki, Nov 19 '21 at 21:32
@podiki I'm sorry but I don't see why, could you be more precise? — bobinthebox, Nov 19 '21 at 22:16
Your definition of $T$ is a $V$ inner product and your definition of $T_{w_k}$ is an $H$ inner product. But when you calculate $T-T_{w_k}$ it's just a $V$ inner product. — podiki, Nov 19 '21 at 22:22

score 1 · Answer 1 · answered Nov 25 '21 at 15:17

No, the solution is sadly incorrect.

One major mistake is in the step $$ (T-T_{w_k})(v) = (w-w_k,v)_V, $$ which is not true because you have defined $T_{w_k}(u) = (w_k,u)_H$ and not $T_{w_k}(u) = (w_k,u)_V$ (the latter would not allow you to show $T_{w_k}\in H^*$).

Note that there is no equality in any sense between $(\cdot,\cdot)_H$ $(\cdot,\cdot)_V$ (only continuity estimates for the induced norms).

Comments on your approach: I do not think your approach is likely to succeed. You start by picking a sequence $w_k\in V$ with $w_k\to w$. There, you might as well choose the constant sequence $w_k=w$. So, in this approach, there is nothing to be gained by considering the sequence $w_k$ (compared to considering just $w$).

A proof of the density is not so simple in my opinion, and some theory can be quite useful. An idea for a proof can be found here.

Prove a density result in the usual Hilbert triple

1 Answers1