# $(u,v)_{H^1} =0 \Rightarrow (u,v)_{L^2} =0$

Let $\Omega \subset \mathbb{R}^d$ be a open set, $L^2(\Omega)$ be the Lebesgue space, and $H^1(\Omega)$ be the Sobolev space.

Does it hold that

$(u,v)_{H^1} =0 \Rightarrow (u,v)_{L^2} =0$?

No, it doesn't hold. For $u,v\in C^1(\Omega)\cap H^1(\Omega)$ we have $(u,v)_{H^1}=(u,v)_{L^2}+(u',v')_{L^2}$. But the dot product can be negative too, so $(u,v)_{H^1}=0$ implies $(u,v)_{L^2}=-(u',v')_{L^2}$, which is not necessarily $0$.
For $d=1$, $\Omega=(0,1)$ we can consider $u(x)=e^{2 x}$ and $v(x)=e^{-\frac12 x}$. We get $$(u,v)_{H^1}=\int_0^1e^{\frac32x}~dx+\int_0^1-e^{\frac32x}~dx=0$$ while $$(u,v)_{L^2}=\int_0^1e^{\frac32x}~dx=\frac23\neq 0.$$ You can also see that the other direction is false too. If $(u,v)_{L^2}=0$ we can't deduce $(u,v)_{H^1}=0$. Here we consider $u(x)=x^2-\frac12$ and $v(x)=x$. We get $$(u,v)_{L^2}=\int_0^1x^3-\frac12 x~dx=\frac14-\frac12\cdot\frac12=0$$ while $$(u,v)_{H^1}=0+\int_0^12x~dx=1\neq 0.$$ All in all, you see that orthogonality in $L^2$ and $H^1$ are not the same.