Orthogonality in terms of two inner products in $W_0^{1,2}(U)$ Let $U$ be an open, bounded, connected set in $\mathbb R^n$.
We define $W_0^{1,2}(U)$ as the closure of $C_c^\infty(U)$ in $W^{1,2}(U)$.
We know $$\langle f,g \rangle_1=\int_U fg+\sum_{i=1}^n \partial_{x_i} f\, \partial_{x_i} g \,dx$$ is an inner product in $W^{1,2}(U)$.
We need the $\int_U fg \, dx$ term to avoid that constant functions have zero norm(induced by the inner product) in $W^{1,2}(U)$.
But we know the unique constant function in $W_0^{1,2}(U)$ is the zero function. We don't need the $\int_U fg \, dx$ term to distinguish zero function from other constant functions(because the only constant function is the zero function). I have shown that $$\langle f,g \rangle_2=\int_U \sum_{i=1}^n \partial_{x_i} f\, \partial_{x_i} g \,dx$$
also defines an inner product in $W_0^{1,2}(U)$.
Now I want to prove that $\forall f,g\in W_0^{1,2}(U)$, if $\langle f,g \rangle_1=0$, then $\langle f,g \rangle_2=0.$ That is, the orthogonality of $\langle \,,\, \rangle_1$ implies the orthogonality of $\langle \,,\, \rangle_2$. But I don't know how to proceed and don't even know if this implication is true. Can someone give me a hint or any advice? Thanks.
 A: Unfortunately the result is not true.
Example: Consider $U=(-\pi,\pi)\subset\mathbb{R}$.  Let $\eta$ be an even bump function that is supported on $[-2,2]$, $1$ on $[-1,1]$ and strictly increasing on $[-2,-1]$.  (For example, $\eta=1_{[-\frac32,\frac32]}\ast\varphi_{1/2}$ where $\varphi_{1/2}(x)=2\varphi(2x)$, $\varphi$ is the standard mollifier).
Pick translates of $\eta$, $f(x)=\eta(x+1)$, $g(x)=\eta(x-1)$ viewed as elements of $C^\infty_c(U)\subset W^{1,2}_0(U)$.  Then $\langle f,g\rangle_2=\int_Uf'g'=0$ since $f'g'$ is the zero function ($\operatorname{supp}f'=[-3,-2]\cup[0,1]$ intersects $\operatorname{supp}g'=[-1,0]\cup[2,3]$ only at $0$).  However, we pick up the contribution $\int_U fg>0$ for $\langle f,g\rangle_1$.

Addendum: This proves $\langle f,g\rangle_2=0\not\Rightarrow\langle f,g\rangle_1=0$.  But that suffices since we can simply choose the pair $f,h:=g-\frac{\langle f,g\rangle_1}{\langle f,f\rangle_1}f$ which will give $\langle f,h\rangle_1=0$ but $\langle f,h\rangle_2=-\langle f,g\rangle_1\langle f,f\rangle_2/\langle f,f\rangle_1<0$.  Or with less computation: in the 2-dimensional subspace spanned by $f,g$, we must have $f^{\perp_1}$ and $f^{\perp_2}$ distinct codimension-1 subspaces (i.e., lines), so there exists some nonzero $h\in f^{\perp_1}$ and we have shown $h$ is not in $f^{\perp_2}$ as the lines are distinct.
