define the Fourier transform and its inverse $$\hat{u}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^n}e^{-ixy}u(x)\,dx$$ $$\check{u}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^n}e^{ixy}u(x)\,dx$$
then there are two relations:
$$\tag{1}(u*v)^{\hat{}}=(2\pi)^{-n/2}\hat{u}\hat{v}$$ $$\tag{2}(uv)^{\hat{}}=\hat{u}*\hat{v}$$
where $*$ means convolution.
but I think there is a contradiction!
apply (1) to $\check{u},\check{v}$: $$(\check{u}*\check{v})^{\hat{}}=(2\pi)^{-n/2}uv$$
hence $$(\check{u}*\check{v})=(2\pi)^{-n/2}(uv)^{\check{}}$$
since $\hat{u}(x)=\check{u}(-x)$
we get : $$(\hat{u}*\hat{v})=(2\pi)^{-n/2}(uv)^{\hat{}}$$
but in $(2)$, the coefficient disappeared. what's the problem?
oh, I found the secret.. it should be $$\tag{1}(u*v)^{\hat{}}=(2\pi)^{n/2}\hat{u}\hat{v}$$ $$\tag{2}(uv)^{\hat{}}=(2\pi)^{-n/2}\hat{u}*\hat{v}$$