What assumptions are necessary for the Fourier transform to exist?

Question

In an article I am reading they apply Fourier transform twice to obtain the following: $$ \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} f(x) e^{- 2 \pi i x \xi} dx d\xi = f(0). $$ I was just wondering what assumptions on $f$ are necessary for this to be valid? Clearly it doesn't hold for all $f$. In particular, is it enough for $f$ to have bounded variation? (This seems to be the only claim they are making about $f$ in the article...) If not I would appreciate if someone could list what would be necessary. Thank you very much.

Answer 1

The basic condition is that $f$ and $\hat f$ are both integrable. Of course that can be difficult to apply, since it's hard to tell whether $\hat f$ is integrable just by looking at $f$. Saying $f$ is a Schwartz function is way more than enough; two weaker conditions that each imply $\hat f$ is integrable are (i) $f,f''\in L^1$ and (ii) $f,f'\in L^2$.

Proof: For (i), note that $\widehat{f''}(\xi)=-\xi^2\hat f(\xi)$. Now $\hat f$ is continuous, so $\int_{-1}^1|\hat f(\xi)|<\infty$. And $\xi^2|\hat f(\xi)|\le||f''||_1$, so $\int_{|\xi|>1}|\hat f(\xi)|\le||f''||_1\int_{|\xi|>1}\xi^{-2}<\infty$.

The proof of (ii) is similar, using the Plancherel theorem and applying Cauchy-Schwarz to $\int_{|\xi|<1}|\hat f|=\int_{|\xi|<1}|\hat f|\cdot 1$ and $\int_{|\xi|>1}|\hat f(\xi)|=\int_{|\xi|>1}|\xi\hat f(\xi)||\xi|^{-1}$.

Answer 2

The value of inner integral is $\widehat{f}(\xi)$, and the value of outer integral is then $(\widehat{f})^{\vee}(0)$, if $f$ is a Schwartz function, then $(\widehat{f})^{\vee}(0)=f(0)$.