I want to prove that $u,v∈H^1 (R^n )$ we have : $∫_{R^n} \frac{∂u}{∂x_i} vdx=-∫_{R^n} \frac{∂v}{∂x_i} udx$
We know that $C_c^∞$$(R^n )$ is dense in $H^1$$(R^n)$, this means that $∃ u_n∈C_c^∞ (R^n )$ such that $u_n→u$ in $H^1 (R^n )$, same for $v$ we have a $v_n$.
Remainder:$u_n→u$ in $H^1 (R^n )$ means that $u_n→u$ in $L^2 (R^n )$ and $\frac{∂u_n}{∂x_i}→\frac{∂u}{∂x_i}$ in $L^2 (R^n )$
Now consider the map: $⟨u,v⟩=∫_{R^n} uv dx$. We know that $⟨*,*⟩$ is a continues operator with respect to the $L^2 (R^n )$ topology, this means that we can interchange the limit of sequences and the integral.
It is trivial to prove:
$∫_{R^n} \frac{∂u_n}{∂x_i} v_mdx=-∫_{R^n} \frac{∂v_m}{∂x_i} u_ndx$
Now we take a limit with respect to “n” and because $⟨*,*⟩$ is continuous with $L^2 (R^n )$ topology and $u_n→u$ in $L^2 (R^n )$ and $\frac{∂u_n}{∂x_i}→\frac{∂u}{∂x_i}$ in $L^2 (R^n )$ we pass the limit inside the integral :
$∫_{R^n} \frac{∂u}{∂x_i} v_mdx=-∫_{R^n} \frac{∂v_m}{∂x_i} udx$
The exact same can be done with a limit with respect to “m” and:
$∫_{R^n} \frac{∂u}{∂x_i} vdx=-∫_{R^n} \frac{∂v}{∂x_i} udx$
Is my aproach correct ?
