Parseval’s relation of Fourier transform

Question

The (Parseval’s relation) is $$\int_{R^n}f(x)\bar{h}(x)=\int_{R^n}\hat{f}(x)\bar{\hat{h}}(x)$$

But does the following relation hold: $$\int_{R^n}f(x)h(x)=\int_{R^n}\hat{f}(x)\hat{h}(x)?$$

If I add another condition: If $h$ satisfies $-\Delta h(x)=L(x)$ where $\int L(x)dx=0$?

Answer 1

No it does not in general. It is simple to see by substituting $h(x) = \bar g(x)$ in the first equation to get $$ \int f(x)g(x) = \int F(x) \bar K(x) $$ where $ K(x) $ is fourier transform of $ \bar g(x)$, which is $ \bar G(-x)$ by property. Therefore, the second equation becomes $$ \int f(x)g(x) = \int F(x) G(-x) $$