# Uniqueness of weak derivatives

Let $\Omega$ be an open set in $\mathbb{R}^n$ and let $f\in L^1_{\text{loc}}(\Omega)$. To show that weak derivatives are unique up to a set of measure zero, suppose $g_{x_i}, \tilde{g}_{x_i}\in L^1_{\text{loc}}(\Omega)$ are two weak derivatives of $f$ with respect to $x_i$. Then, $$\int_{\Omega}f\varphi_{x_i}\, dx=-\int_{\Omega}g_{x_i}\varphi\, dx$$ and $$\int_{\Omega}f\varphi_{x_i}\, dx=-\int_{\Omega}\tilde{g}_{x_i}\varphi\, dx$$ for any $\varphi\in C^{\infty}_c(\Omega)$. So we have $$\int_{\Omega}(g_{x_i}-\tilde{g}_{x_i})\varphi\,dx=0.$$ From here, how can I conclude $g_{x_i}=\tilde{g}_{x_i}$ almost everywhere on $\Omega$? If someone knows the answer, please let me know.

• Are the weak derivatives assumed to exist only as generalized functions or as locally integrable functions? Your notations seem to indicate that they are locally integrable. In that case what you are asking is: $h$ locally integrable, $\int h\phi =0$ for all $C^{\infty}$ functions $\phi$ with compact support then $h=0$ a.e.. Is this what you are asking? Sorry if I have not read your question properly. – Kavi Rama Murthy Apr 4 '18 at 5:31
• By approximating $1_B$ you can show that it is true for indicator functions of open balls, then open sets, then Borel sets and then Lebesque measurable sets. Then choose the indicator function of the set where $g_{x_i}-\tilde{g}_{x_i} >0$. – copper.hat Apr 4 '18 at 5:34
• @KaviRamaMurthy Yes two weak derivatives are assumed to be locally integrable. Sorry I did not specify. – gladimetcampbells Apr 4 '18 at 5:37

Yes. We are to prove that if $$f\in L^1_{\rm{loc}}(\Omega)$$ satisfies $$\displaystyle\int_{\Omega}f\varphi=0$$ for all $$\varphi\in C_{0}^{\infty}(\Omega)$$, then $$f=0$$ a.e. on $$\Omega$$.
Let $$K$$ be any compact subset of $$\Omega$$, and take $$\psi\in C_{0}^{\infty}(\Omega)$$ such that $$\psi=1$$ on $$K$$. Define a function $$f_\psi$$ on $$\mathbb{R}^n$$ as $$\begin{equation*} f_{\psi}(x)= \begin{cases} 0 & x\notin\Omega, \\ f(x)\psi(x) & x\in\Omega. \end{cases} \end{equation*}$$ which extends $$f$$ canonically to the whole $${\bf{R}}^{n}$$. Then $$f_{\psi}\in L^{1}({\bf{R}}^{n})$$. Pick a mollifier $$\varphi\in C_{0}^{\infty}({\bf{R}}^{n})$$. Note that $$\varphi_{\epsilon}\ast f_{\psi}\rightarrow f_{\psi}$$ in $$L^{1}_{\rm{loc}}({\mathbb{R}}^{n})$$ as $$\epsilon\to 0$$. We have \begin{align*} \varphi_{\epsilon}\ast f_{\psi}(x)=\int f(y)\psi(y)\varphi_{\epsilon}(x-y)dy, \end{align*} and note that for a fixed $$x$$, the map $$y\mapsto\psi(y)\varphi_{\epsilon}(x-y)\in C_{0}^{\infty}(\Omega)$$, so $$\varphi_{\epsilon}\ast f_{\psi}=0$$ by hypothesis. As $$\epsilon\to 0$$, we have $$\varphi_{\epsilon}\ast f_{\psi}\to f_{\psi}$$ and thus $$f_{\psi}=0$$ a.e. on $$\Omega$$
Hence $$f=0$$ a.e. on $$K$$. Since $$K$$ is arbitrary compact subset of $$\Omega$$, we have $$f=0$$ a.e. on $$\Omega$$.
• Locally integrable for $f$ is assumed. As @Kavi Rama Murthy indicated there. – user284331 Apr 4 '18 at 5:33