The function is 0 almost everywhere

[evinda]

Hello! (Wave)

Let $ u \in L^1_{\text{loc}}(\Omega) $ and the weak derivatives of first class satisfy $ D_{x_i} u(x)=0 , i=1, \dots, n $ almost everywhere in $ \Omega $.
I want to show that $ u=0 $ almost everywhere.

I have thought the following:

$ 0= \int_{B(x, \epsilon)} D_{x_i}u(x) \psi_{\epsilon}(x-y) dy \overset{\epsilon \to 0}{\to} D_{x_i} u(x) $

where $ x $ is a Lebesgue point and $ \psi_{\epsilon}(x)=\epsilon^{-n }\psi{\left( \frac{x}{\epsilon}\right)}, \epsilon>0 $

where $ \psi \in C_C^{\infty}(\mathbb{R}^n), \psi>0, \int \psi(x) dx=1 $.

Is it right so far? Can we use this in order to show that $u=0$ almost everywhere?Does it hold that $ \int_{B(x, \epsilon)} D_{x_i}u(x) \psi_{\epsilon}(x-y) dy=- \int_{B(x, \epsilon)} u(x) \psi_{\epsilon}'(x-y) dy $ ?

 

[Jhoneine]

Yes, it does hold. Using integration by parts, we have $$ \int_{B(x, \epsilon)} D_{x_i}u(x) \psi_{\epsilon}(x-y) dy = - \int_{B(x, \epsilon)} u(x) \frac{\partial}{\partial x_i}\psi_{\epsilon}(x-y) dy = - \int_{B(x, \epsilon)} u(x) \psi_{\epsilon}'(x-y) dy $$and since $ \psi_{\epsilon}'(x-y) \to 0 $ as $ \epsilon \to 0 $, we have that $ D_{x_i}u(x) = \lim_{\epsilon \to 0} \int_{B(x, \epsilon)} D_{x_i}u(x) \psi_{\epsilon}(x-y) dy = 0$.Since $ D_{x_i}u(x) = 0 $ for all $ i=1,\dots, n $ and all Lebesgue points of $ u $, this means that $ u $ must be zero almost everywhere.

 

[math771]

Yes, your approach is correct so far. To show that $u=0$ almost everywhere, we can use the fact that $D_{x_i}u(x)=0$ almost everywhere. This means that for almost every $x \in \Omega$, we have $D_{x_i}u(x)=0$ for all $i=1,\dots,n$.

Now, let $x$ be a Lebesgue point of $u$, which means that the limit in your first equation exists. Then, as $\epsilon \to 0$, we have
$$0=\lim_{\epsilon \to 0}\int_{B(x,\epsilon)} D_{x_i}u(x)\psi_{\epsilon}(x-y)dy=\int_{\mathbb{R}^n}D_{x_i}u(x)\psi(x)dx=0$$
where we have used the fact that $\psi_{\epsilon}(x-y) \to \psi(x)$ as $\epsilon \to 0$ and the definition of a Lebesgue point.

Since this holds for all $i=1,\dots,n$, we have
$$0=\int_{\mathbb{R}^n}D_{x_i}u(x)\psi(x)dx=0$$
for all $i=1,\dots,n$. But this is just the definition of a weak derivative, so we have shown that $u$ has weak derivatives of first class equal to $0$ everywhere in $\Omega$. This implies that $u$ is constant almost everywhere in $\Omega$, and since it is in $L^1_{\text{loc}}(\Omega)$, it must be $0$ almost everywhere.

As for your second question, yes, it does hold that
$$\int_{B(x,\epsilon)} D_{x_i}u(x)\psi_{\epsilon}(x-y)dy=-\int_{B(x,\epsilon)} u(x)\psi_{\epsilon}'(x-y)dy$$
This can be seen by integrating by parts and using the fact that $D_{x_i}u(x)=0$ almost everywhere.