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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.

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  • $\begingroup$ Yes, see here: math.stackexchange.com/questions/2681211/…. $\endgroup$ Oct 27, 2021 at 14:36
  • $\begingroup$ "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?" obviously yes. Let $f=0$. Then $T_f=0=T$, and $f=0$ a.e. .(The version of the question that has some content is this: "If $T_f=0$ does it follow that $f=0$ ae?") $\endgroup$ Oct 27, 2021 at 22:38

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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 $f-g=0$.

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  • $\begingroup$ I rephrased my post more accurately. Thanks $\endgroup$
    – C.G.
    Oct 27, 2021 at 16:58
  • $\begingroup$ @C.G. Somehow you missed my point - my "surely that's not exactly what you meant, because..." applies to the revied version of the question as well. And for exactly the same reason. $\endgroup$ Oct 27, 2021 at 22:35
  • $\begingroup$ The question I asked was what I meant. In your answer, you assume that $f\in L_{\mathbb{loc}}^1(\mathbb{R}^n)$ and I agree then that the result is straightforward. My question is what happens when T=S in the sense of distributions knowing only that S can be identified to a function $g\in L_{loc}^1$. $\endgroup$
    – C.G.
    Oct 28, 2021 at 5:55
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    $\begingroup$ What happens in that case is yes, you can obviously identify $T$ with $f$ by letting $f=g$. $\endgroup$ Oct 28, 2021 at 9:45

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