Equality in the sense of distributions

Question

Let $T$ and $T_g$ be two distributions belonging to $\mathcal{D}'(\mathbb{R}^n) \times \mathcal{D}'(\mathbb{R}^n)$. We assume that $T_g$ can be identified to a function $g\in L^1_{\mathbb{loc}}(\mathbb{R}^n)$ (using the fact that the mapping $h\in L^1_{\mathbb{loc}}(\mathbb{R}^n) \mapsto T_h\in \mathcal{D}'(\mathbb{R}^n)$ defined by $\forall \varphi\in \mathcal{D}(\mathbb{R}^n)$, $\langle T_h,\varphi\rangle=\displaystyle{\int_{\mathbb{R}^n}}\,h\,\varphi\,dx$ is an injection). If $T=T_g$, can we conclude that the distribution $T$ can be identified to a function $f\in L^1_{\mathbb{loc}}$ equal to $g$ almost everywhere? Also, if $T=0$ in the sense of distributions, can we conclude that $T$ can be identified to a function equal to zero almost everywhere? I apologize if the questions are obvious. Best regards.

Answer 1

I doubt that "If $T_f=T_g$, can we conclude that the distribution $T_f$ can be identified to a function $f\in L^1_{loc}$ equal to $g$ almost everywhere?" is exactly what you meant to ask, because it's completely trivial: Yes, let $f=g$; then $T_f=T_g$ and $f=g$ almost everywhere.

The "real" question is this: "If $T_f=T_g$ does it follow that $f=g$ ae?" (no, that's not the same question! In my version $f$ is at least implicitly given, while your version asks whether there exists $f$ with certain properties.) The answer is again yes: We know that $$\int (f-g)\phi=0\quad(\phi\in\mathcal D).$$Suppose that $\phi\in C_c(\Bbb R^d)$. there exists a sequence $(\phi_n)\subset\mathcal D'$ supported in a fixed compact set and tending to $\phi$ uniformly; hence $\int(f-g)\phi=0$. Since this holds for every $\phi\in C_c$ it's well known that it follows that $f-g=0$.