If we choose to define formally the FT of a function by
$$\hat f(\xi) = {1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi
x}dx$$
rather than
$$\hat f(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi x}dx$$
then formulas change. for example
$$\widehat{f \star g} (\xi) = \sqrt{2\pi} \hat f(\xi) \hat g(\xi)$$
rather than
$$\widehat{f \star g} (\xi) = \hat f(\xi) \hat g(\xi)$$
my question is do theorems change also? is the fourier transform still an isometry from $L^1(R) \cap L^2(R)$ to $L^2(R)$? or is their some constant that will pop up?
