Take the Sobolev space $$H^1(T)=\left\{f \in L^2(T) ~|~ f' \in L^2(T)\right\}$$ where $T$ is the 1-torus (that is a circle) and $f'$ the weak derivative. Take then a function $v\in H^1(T)$ and define for every $f,g\in H^1(T)$ the $L^2(T)$-inner-product $$(f,g)_{L^2(T)}:= \int_T f\cdot g ~d\mu $$ and the $H^1(T)$-inner-product $$(f,g)_{H^1(T)}:= (f,g)_{L^2(T)} + (f',g')_{L^2(T)} $$ where $d\mu$ is the Lebesgue measure and $f'$ and $g'$ are the weak derivatives.
The question is the following: it is possible to find a function $w\in H^1(T)$ that is orthogonal to $v$ in the sense of the $L^2(T)$-inner-product, but it is not orthogonal to $v$ in the sense of the $H^1(T)$-inner-product? In case, can you exhibit a counterexample or prove that it is not possible?
Thank you
Suppose $v \in H^1(T)$ is not essentially constant and $\int_T v \neq 0$. Define $a=-\frac{\int_T v^2}{\int_T v}$. Then $v$ and $a+v$ are orthogonal in $L^2$ (check it) but not in $H^1$ (the derivatives are equal everywhere, so the inner product is the integral of a nonnegative function which is nonzero on a set of positive measure).
The first assumption is essential: among essentially constant functions the $H^1$ inner product reduces to the $L^2$ inner product. I don't know whether the second assumption is essential (though I doubt it).