# If a subspace is dense in a reflexive Banach space, then “it” is dense in the dual?

It is well-known that if a Banach space $X$ is reflexive and separable, then its dual $X'$ is separable. My question is more concrete, but for clarity, I am going to state a few questions going from most general to most concrete:

1. Suppose it is known that the Banach space $X$ is reflexive, and a subspace $Z$ of $X$ is dense in $X$. Let $T:Z\rightarrow X'$ be an arbitrary injective embedding. Does it follow that $T(Z)$ is dense in $X'$? Answer: No (see Nate's comment).

2. Suppose $X$ is a Hilbert space, and $Z$ is a subspace of $X$ which is dense in $X$. Do not identify $X$ with its dual. Furthermore, suppose that there exists a map $T:X\rightarrow X'$ such that the elements of $T(Z)$ coincide with a subset of those of $Z$. Does it follow that $T(Z)$ is dense in $X'$?

3. Suppose $X$ is a Hilbert space whose elements are measurable functions and such that the elements of a subspace $Z$ of $X$ coincide with those of $C_c^{\infty}$, and $Z$ is dense in $X$. Do not identify $X$ with its dual. Furthermore, suppose that the elements of a subspace $Z'$ of $X'$ coincide with those of $C_c^{\infty}$. Does it follow that $Z'$ is dense in $X'$?

Note that in the concrete case of question 3 with $X=H^1=W^{1,2}(\mathbb R^n)$, the question has been resolved with answer "yes", by reuns below.

I am most interested in the answer to question 3, in the general setting.

• For the general case, how is $C_c^\infty$ identified with a subspace of $X'$? – Pedro Aug 24 '17 at 3:31
• I think that the answer depends essentially on how $C_c^\infty$ is seen as a subspace of $X'$. In the general case, the assumption "$C_c^{\infty}$ is a subspace of $X'$" means only that an injective linear map $T:C_c^{\infty}(\mathbb R^n)\to X'$ is given (under the usual identification in the Hilbert space case, $T$ is an isometry). You want to know whether $T(C_c^\infty)$ is dense in $X'$. I think that only linearity and injectivity are not enough to prove it (in the Hilbert space case under the usual identification, $T$ is also defined on the entire space, bijective and continuous). – Pedro Aug 24 '17 at 6:36
• As $X$ is arbitrary and $C_c^\infty$ is seen as subspace of $X'$ in no specific way, I cannot see what is special about $C_c^\infty$. So, I think your question can be written as: given a reflexive space $X$, a dense subspace $Z\subset X$ and an injective linear map $T:Z\to X'$, is it true that $T(Z)$ is dense in $X'$? What about in addition $X$ is Hilbert? – Pedro Aug 24 '17 at 6:57
• @Pedro , well I am primarily interested in spaces where elements are measurable functions, so the elements of $T(Z)$ should really coincide with those of $Z$ in the sense of measurable functions. But, it is true that I do not want $T=\psi$. Take the concrete case of $X=W^{1,2}(\mathbb R^n)$. Then $$X\subsetneq L^2(\mathbb R^n)\subsetneq X',$$ and it is clear that the smooth compactly supported functions form subspaces in all three spaces in the above chain. – Lentes Aug 24 '17 at 23:57
• The most usual setting for this sort of thing is a Gelfand triple; you have a reflexive Banach space $X$ and a Hilbert space $H$ with a continuous injection $i : X \to H$ whose image is dense. Then you can show the adjoint $i' : H \to X'$ is also a continous injection whose image is dense. Thus if $E$ is dense in $X$, then $i'(i(E))$ is dense in $X'$. – Nate Eldredge Aug 25 '17 at 21:47

## 1 Answer

For $W^{1,2}$ you can try something like this :

As $C^\infty_c$ is dense in the distributions, it is more or less obvious $C^\infty_c$ is dense in the weak dual of $H^1 = W^{1,2}$ the Hilbert space with norm $\|f\|_{H^1}^2 = \|f\|_{L^2}^2+\|f'\|_{L^2}^2$.

So the question is if it is also dense in the strong dual.

With the Fourier transform $$\|f\|_{H^1}^2 = \int_{-\infty }^\infty |\widehat{f}(\xi)|^2d\xi +\int_{-\infty }^\infty |\widehat{f}(\xi)|^24 \pi^2 \xi^2 d\xi = \|\widehat{f} \sqrt{1+4 \pi \xi^2}\|^2_{L^2}$$ Thus its strong dual is $H^{-1}$ the Hilbert space with norm $\|g\|_{H^{-1}}^2 = \|\frac{\textstyle\widehat{g} }{\sqrt{1+4 \pi \xi^2}}\|^2_{L^2}$. It is not hard to check the Schwartz space is dense on the Fourier side for the $\|\frac{. }{\sqrt{1+4 \pi \xi^2}}\|^2_{L^2}$ norm, and as the Schwartz space is its own Fourier transform, the Schwartz space is dense in $H^{-1}$, and so is $C^\infty_c$.

• Thanks for this, reuns; this is very helpful. Yes, I do mean denseness in the dual space in the strong topology. The abstract question, even in the Hilbert space setting, is still open, so I'll leave the question open for now. – Lentes Aug 25 '17 at 0:52
• reuns and @Lentes My conclusion is that $C_c^\infty(\mathbb R^n)$ is not dense in ($W^{1,2}(\mathbb R^n))'$ (see my post). What is the mistake? – Pedro Aug 25 '17 at 4:16
• @Pedro Are you saying if $T$ is continuous then $T(X)$ is a closed/complete subspace of $X'$ ? For $u \in l^2$ define $\|u\|_B^2 = \sum_n |u_n|^2 e^{-n^2}$ then $l^2$ is dense in $B$ and the identity $\iota : l^2 \to B$ is continuous but $\iota(l^2) \ne B$ – reuns Aug 25 '17 at 4:43
• @reuns I was wrong about that (in fact, in general only one side of the inclusion is valid). I deleted the post. Thanks. – Pedro Aug 25 '17 at 5:05
• Yeah; there is actually a different argument to conclude that $C_c^{\infty}$ is dense in $H^{-1}$ (with the added nicety that the argument works if we replace $\mathbb R^n$ by any open set $\Omega$). The argument is this: if $f\in H^{-1}$, then by Riesz Representation there exists a unique $u\in H^1$ such that $$-\Delta u+u =f$$ in the weak sense. The above actually defines an invertible, isometric mapping $\psi: H^1\rightarrow H^{-1}$. It is clear that $\psi(C_c^{\infty})\subseteq C_c^{\infty}$, and since $\psi(C_c^{\infty})$ is dense in $H^{-1}$, so too must be $C_c^{\infty}$. – Lentes Aug 25 '17 at 18:48