# Equality in the sense of distributions

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.

• Yes, see here: math.stackexchange.com/questions/2681211/…. Oct 27, 2021 at 14:36
• "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?") Oct 27, 2021 at 22:38

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 $$f-g=0$$.
• 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 $g\in L_{loc}^1$.
• What happens in that case is yes, you can obviously identify $T$ with $f$ by letting $f=g$. Oct 28, 2021 at 9:45