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?

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.