Interpretation of lemma: $f\in L^{1}_{loc}(\Omega)\ st\ \int f\phi=0\: \forall\phi\in\mathcal{D}(\Omega)\ \Rightarrow f=0\ in\ L^{1}_{loc}(\Omega)$

Question

As in the title, consider the following lemma in the theory of distributions:
$$ f\in L^{1}_{loc}(\Omega)\;\;\text{ s.t. }\int f\phi=0\quad \forall\phi\in\mathcal{D}(\Omega)\ \implies f=0\;\text{ in }\; L^{1}_{loc}(\Omega) $$
where $\mathcal{D}(\Omega)$ is the space of compactly supported $C^{\infty}$ functions, also known as test functions.

The proof I know is based on mollifiers and convolution: the result follows by the uniform convergence of the regularized $f_\epsilon=f\ast\rho_\epsilon$ to $f$, as $\epsilon\rightarrow 0$ where $\rho_\epsilon$ are a family of mollifiers.

Now, we can see the same result written as $\langle f,\phi\rangle=0\; \forall\phi\in\mathcal{D}(\Omega)\ \Rightarrow\ f=0\ in\ L^1_{loc}(\Omega)$ where we view $f$ as an element of the dual $\mathcal{D}'(\Omega)$. This may be interpreted as $f\in\mathcal{D}(\Omega)^{\perp}\subset\mathcal{D}'(\Omega)\ \Rightarrow f=0$.
If we where in a (seprable) Hilbert space, where the inner product allows us to 'internalize' the notion of dual (through Riesz theorem) and as a consequence that of orthogonal (as the most natural one) this consition is known to be necessary and sufficient for the set we are treating to be a orthogonal basis.

I wonder if:

• we can have an analogous in this case (maybe through the notion of density of $\mathcal{D}(\Omega)$ in $L^1$ ?) where the setting is that of general Topological Vector Spaces
• It is in general possible to define and charachterize a subspace $M\subset E$ such that $f\in E\ st\ \langle f,\phi\rangle =0\ \forall\phi\in M\ \Rightarrow f=0\ \in E'$
• If yes, can we in some sense think of M as a ''basis''? More in general are there relevant generalization of the concept in arbitary TVS?

Answer 1

You can use in general this fact:

For each space $Y$ the diagonal $\Delta$ is closed if and only if $Y$ is Hausdorff. So you have the following criteria to estabilish when two continuos function $f,g: X\to Y$ must be equal:

If $f$ and $g$ coincides on a dense set of $X$, then $f=g$.

In our case we have that the linear map

$\langle f, \cdot \rangle: V\to \mathbb{R}$

is a countinuos map with respect the topology induced by the metric induced by the scalar product on $V$ so if the map is zero on a subset $A$ dense on $V$ then the map is zero on all $V$.

In your first case you have that $D(\Omega)$ is dense in $L_{loc}^1(\Omega)$ by theory of mollifiers, so if the scalar product is zero on $D(\Omega)$ then is zero on all $L_{loc}^1(\Omega)$

So if $M$ is dense on $E$ then you have your result. I don’t know if it is also a necessary condition on $M$ to have your property.