MHB 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.