Prove that different functions in $L^1_{\operatorname{loc}}(\Omega)$ define different distributions in $\mathcal D’(\Omega)$

Question

Let $\Omega \subset R^n$ be an open set, for $u\in L_{\rm loc}^1(\Omega)$, the mapping $\varphi\in\mathcal{D}(\Omega)\mapsto\int_\Omega u(x)\varphi(x)\mathrm{d}x$ defines a distribution.

Let $\mathcal{D}'(\Omega)$ be the set of all the distributions defined on $\mathcal{D}(\Omega)$, I want to show that $\mathcal{l}:L_{\rm loc}^1(\Omega)\rightarrow \mathcal{D}'(\Omega)$ is injective.

Here is a proof I have read recently:

We shall only prove $\mathcal{l}^{-1}(0)=\{0\}$. Since $\mathcal{D}(\Omega)$ is dense in $L^1(\Omega)$, if $\int_\Omega u(x)\varphi(x)\mathrm{d}x=0, \forall\varphi\in\mathcal{D}(\Omega)$, then we must have $u\in [L^1(\Omega)]'\sim L^\infty(\Omega)$ (???). It follows that $u=0$ in $L^\infty(\Omega)$, i.e. $u\stackrel{a.e}{=}0$.

Please help by explaining how is (???) obtained, in other words, how can I tell that $\varphi \mapsto\int_\Omega u(x)\varphi(x)\mathrm{d}x$ is a $\textbf{bounded}$ linear functional on $L^1(\Omega)$.

(I have learned another proof, which is rigorous, but way more complicated than this one. I suspect there may be mistakes in the proof above.)

Answer 1

The proof you wrote is intuitive, but it is simply a mess from a formal point of view. The fact that $\mathcal D(\Omega)$ is dense in $L^1$ implies that the above functional $$ T:\mathcal D(\Omega)\to \mathbb R $$ $$ \varphi\mapsto \int \varphi u $$ uniquely extends on a bounded linear functional $\overline T$ in $(L^1(\Omega))’$,

$$ \overline T:L^1(\Omega)\to \mathbb R $$

(extension is possible because bounded linear functions are uniformly continuous, check this link. Alternatively, use Hahn Banach to extend the linear functional on all $L^1$ functions and the density of test functions to show the extension is unique).

Since $(L^1)’$ is identified with $L^\infty$, you know that there exists a unique function $\overline u\in L^\infty$ such that

$$ \overline T:\varphi\mapsto \int \varphi \overline u $$

(actually, in this case $\overline u=0$). The problem is, you do not know a priori that $u=\overline u$ a.e.. To show this, you should consider $w:=u-\overline u$ and prove that $w\equiv 0$ from the fact that $\int \varphi w=0$ for $\varphi\in \mathcal D(\Omega)$ which is just the original problem, so you can see the above proof is not solid, or at least not complete.

One way to conclude from there, actually, would be to show that $u$ is in $L^\infty$ (since we said that the function $\overline u$ is unique in $L^\infty$, and if $u\in L^\infty$, the integral $\int \varphi u$ is well-defined for any $\varphi\in L^1$). To show that $u$ is in $L^\infty$, you need to use the property stating that for a general $u\in L^1_{loc}$, $$ \|u\|_{L^\infty}=\sup_{0\neq\phi\in A}\frac{|\int \phi u|}{\|\phi\|_{L^1}} \qquad (1)$$ whenever the norm on the left hand side is finite or infinite, and where $A$ is any vector space which is dense in $L^1$ (in our case, $A=\mathcal D(\Omega)$). If you take this property for granted, then you would be able to say that $u\in L^\infty$ and that in fact $\overline T$ is defined as $$ \overline T:\varphi\mapsto \int \varphi u. $$ But if you rely on $(1)$, all the previous part of the proof is pointless, you don’t really need to rely on anything else, since from $(1)$ you obtain that $\|u\|_{L^\infty}=0$, that is, $u\equiv 0$ a.e..

I remark that, even if you manage to make the above proof complete, you still miss the crucial part of the proof and hide the true uniqueness part in previously known results (how do you know that $\overline u$ in the definition of $\overline T$ is unique in the first place if you don’t rely on the duality between $L^1$ and $L^\infty$??).

This proof simply relies too much on the abstract functional-analytic tools, without wanting to dig deeper in the details of measure theory. This is how an experienced mathematician would think, because he is sure he can fill the remaining details when needed, but it is not a proof I would teach to a student. It simply takes too much effort to make the proof rigorous, and in my opinion one should look for a proof that relies on more elementaly tools, even though it might look longer. This is probably not what you wanted to hear, but I hope the answer helps you understanding this “proof”.