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 $\Omega$?

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    $\begingroup$ 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. $\endgroup$ Apr 4, 2018 at 5:31
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    $\begingroup$ 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 $g_{x_i}-\tilde{g}_{x_i} >0$. $\endgroup$
    – copper.hat
    Apr 4, 2018 at 5:34
  • $\begingroup$ @KaviRamaMurthy Yes two weak derivatives are assumed to be locally integrable. Sorry I did not specify. $\endgroup$
    – Koda
    Apr 4, 2018 at 5:37

1 Answer 1


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 $\Omega$.

  • $\begingroup$ Locally integrable for $f$ is assumed. As @Kavi Rama Murthy indicated there. $\endgroup$
    – user284331
    Apr 4, 2018 at 5:33
  • $\begingroup$ Thank you so much! $\endgroup$
    – Koda
    Apr 4, 2018 at 5:40
  • $\begingroup$ Just a comment: the lemma proved in this reply is known as "the fundamental lemma of the calculus of variations", since it is the crucial step is obtained the Euler-Lagrange equation for the minimizer of a functional. $\endgroup$ Oct 18, 2019 at 14:35
  • $\begingroup$ Why do you need to canonically extend $f$ to the whole $\mathbf R^n$ ? Doesn't the proof work the same if you directly consider $f$ and a mollifier $\varphi \in C_0^\infty(\Omega)$ ? $\endgroup$ May 2, 2023 at 9:41
  • $\begingroup$ @StratosFair, for how do you interpret $\int_{\mathbb{R}^{n}}f(y)\varphi(x-y)dy$ for the variable $y\in\Omega^{c}$ but $f$ is not defined on $\Omega^{c}$? $\endgroup$
    – user284331
    May 2, 2023 at 12:39

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