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I have recently read the proof of the following theorem, i.e.,

Theorem Let $I$ be an open subset of $\mathbb{R}$. Let $f \in L^1_{\mathrm{loc}} (I)$ such that $\int_I f \varphi' =0$ for all $\varphi \in C^\infty_c (I)$. Then $f$ is constant a.e. in $I$.

Proof We fix a function $\psi \in C^\infty_c (I)$ such that $\int_I \psi = 1$. Let $w \in C^\infty_c (I)$ and $h := w - ( \int_I w) \psi$. Clearly, $h \in C^\infty_c (I)$. We define $\varphi \in C^\infty (I)$ by $\varphi (x) := \int_{-\infty}^x h(s) \, \mathrm d s$. By FTC, $\varphi' = h$. Because $\int_I h =0$, we get $\varphi \in C^\infty_c (I)$. Then $$ \begin{align*} 0 &= \int_I f \varphi' \\ &= \int_I f h \\ &= \int_I f \left ( w - \left ( \int_I w \right ) \psi \right) \\ &= \int_I \left ( f - \int_I f \psi \right) w. \end{align*} $$ Because $w$ is arbitrary in $C^\infty_c (I)$, we get $f = \int_I f \psi$ a.e. on $I$.

I would like to ask if above theorem holds in higher dimension, i.e.,

Let $\Omega$ be an open subset of $\mathbb{R}^d$. Let $f \in L^1_{\mathrm{loc}} (\Omega)$ such that $\int_\Omega f \nabla \varphi =0$ for all $\varphi \in C^\infty_c (\Omega)$. Then $f$ is constant a.e. in $\Omega$.

Thank you so much for your elaboration!

Akira
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  • Does $f(\nabla\varphi)$ mean $f\circ \nabla\varphi$ or $f\nabla\varphi$? – kieransquared Nov 05 '23 at 22:39
  • @kieransquared i meant the latter. – Akira Nov 05 '23 at 22:52
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    Both statements are false if $I$, resp. $\Omega$ are not connected. In the proof of the Theorem above, it is implicitly assumed that $I$ is an interval, otherwise the proof cannot work. – Lorenzo Pompili Nov 06 '23 at 11:33
  • @LorenzoPompili Could you explain where we use the connectedness of $I$? – Akira Nov 06 '23 at 11:36
  • If you assume that $\Omega$ is connected, then yes, the analogous theorem in higher dimensions holds. I cannot think of a reference, but it is definitely a classical result. A possible proof goes like that: you first assume that $\Omega$ is a ball, in that case you can e.g. use a distributional argument, like in this answer: https://math.stackexchange.com/a/4191922/884561 . Then, for general $\Omega$, you know that $f$ is a.e. constant in any ball contained in $\Omega$, and you need to use connectedness of $\Omega$ to prove that the constant does not depend on the ball you choose. – Lorenzo Pompili Nov 06 '23 at 11:40
  • @LorenzoPompili in above proof for the case of $1$D, I could not see where the connectedness of $I$ is used. Could you elaborate on this point? – Akira Nov 06 '23 at 11:43
  • Well, if $I$ is not an interval, then in general the support of $\varphi$ is not contained in $I$. An example: consider $I$ as the union of two disjoint open intervals, and choose $w\in C^\infty_c(I)$ s.t. the integral of $w$ on $I$ is zero (so that $h=w$), and such that the integral of $w$ in any of the two intervals is non-zero. But in the proof above you wrote that $\varphi\in C^\infty_c(I)$, so you are indeed using that $I$ is an interval. – Lorenzo Pompili Nov 06 '23 at 11:45
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    P.s. Showing that the above theorem is wrong when $I$ is not an interval is not difficult. Consider again $I$ as the union of two disjoint intervals. Then, any function which is constant in any of the two intervals, but such that the constant is not the same on the two intervals, satisfies the hypotheses of the theorem (and is not equal to a constant a.e.). At most, for a general open set $I$, you can prove that $f$ is locally a.e. constant. – Lorenzo Pompili Nov 06 '23 at 11:54
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    @LorenzoPompili I got it. Thank you so much for your help! – Akira Nov 06 '23 at 11:56

1 Answers1

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I would try it with smoothing by convolutions. Here is a sketch:

Let $\rho\in C_{c}^{\infty}\left( B\right) $ be a mollifier with $\rho\geq0$, $\rho \left( -x\right) =\rho \left( x\right) $ and $\int_{B}\rho\,\mathrm{d}x=1$, where $B$ is the unit ball around the origin. Set $\rho _{k}\left( x\right) =k^{n}\rho\left( kx\right) $. Then the convolution $\rho_{k}\ast f$ is smooth and $$ \rho_{k}\ast f\rightarrow f\quad\text{in }L_{\mathrm{loc}}^{1}\left( \Omega\right) , $$ in particular, $$ \left( \rho_{k}\ast f\right) \left( x\right) \rightarrow f\left( x\right) \quad\text{for a.e. }x\in\Omega. $$

So $$ 0=\int_{\Omega}f\nabla\left( \rho_{k}\ast\varphi\right) \,\mathrm{d}% x=\int_{\Omega}f\left( \rho_{k}\ast\nabla\varphi\right) \,\mathrm{d}% x=\int_{\Omega}\left( \rho_{k}\ast f\right) \nabla\varphi\,\mathrm{d}% x=-\int_{\Omega}\nabla\left( \rho_{k}\ast f\right) \varphi\,\mathrm{d}x $$ for all $\varphi\in C_{c}^{\infty}\left( \Omega\right) $, implying that $$ \nabla\left( \rho_{k}\ast f\right) =0\quad\Rightarrow\quad\rho_{k}\ast f=\mathrm{const}\quad\text{if }\Omega\text{ is connected.}% $$ By the above, also $$ f\left( x\right) =\mathrm{const}\quad\text{for a.e. }x\in\Omega. $$

Another proof uses the Sobolev space $W^{1,1}$ and the Poincaré inequality $$ \int_{B_{R}}\left\vert f-f_{B_{R}}\right\vert \,\mathrm{d}x\leq c\left( d\right) R\int_{B_{R}}\left\vert \nabla f\right\vert \,\mathrm{d}x $$ for $f\in W^{1,1}\left( B_{R}\right) $, where $B_{R}$ is a ball of radius $R$. Under your hypothesis, $$ f\in W_{\mathrm{loc}}^{1,1}\left( \Omega\right) \quad\text{with}\quad\nabla f=0, $$ so that $f=\mathrm{const}=f_{B_{R}}$ on any ball $B_{R}\Subset\Omega$.

As a reference on smoothing by convolutions and on Sobolev spaces, I recommend Gilbarg & Trudinger, Elliptic Partial Differential Equations of 2nd Order, Chapter 7.

Christoph
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