Question about the Fourier Inversion Formula

Question

We have $$\hat{f}(\xi)=\mathcal{F}f(\xi):= \int_{-\infty}^{\infty}f(x)e^{-2\pi i\xi x}dx,$$ with $f\in L^{1}$, and the Fourier inversion formula says that $$f(x)=\int_{-\infty}^{\infty}\hat{f}(\xi)e^{2\pi i\xi x}d\xi.$$ My question is: do we need that $\hat{f}\in L^{1}$, in order to the integral above converges and inversion formula holds, or $f\in L^{1}$ implies somehow $\hat{f}\in L^{1}$? Thanks.

Answer 1

You certainly need to assume that $\hat f\in L^1$ to apply the inversion formula as written; no, that does not follow from assuming $f\in L^1$. It's been pointed out that $f.f'.f''\in L^1(\mathbb R)$ imply $\hat f\in L^1$; you might also note that $f,f'\in L^2(\mathbb R)$ imply $\hat f\in L^1$.

Even if $\hat f\notin L^1$ you can still recover $f$ from $\hat f$ by using a "summability method" on the integral. For example, if $f\in L^1(\mathbb R)$ it follows that $$f(x)=\lim_{y\to0^+}\int_{-\infty}^\infty e^{-y|\xi|}\hat f(\xi)e^{2\pi i\xi x}\,d\xi$$for almost every $x$.

Answer 2

No, it doesn't imply absolute integrability. Yes, both $f$ and $\hat f$ need to be $L^1$. However, if for example $f\in C^2(\mathbb R)$ and $f',f''\in L^1$, then $$\mathcal F [f''](\xi)=(-2\pi i \xi)^2 \hat f (\xi)$$ is a bounded continuous function, and for this to be true $\hat f$ must be bounded at least by something like $$|\hat f(\xi)|\leq \frac{C}{\xi^2+1},$$

so $\hat f \in L^1$. So only marginal smoothness assumptions are necessary for the inversion theorem to hold.

Answer 3

If you want to work in a classical pointwise way, the inversion is a Cauchy Principle Value integral $$ f(x)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\hat{f}(\xi)e^{2\pi i\xi x}d\xi . $$ If $f$ is absolutely integrable and satisfies some condition such as having left- and right-hand derivatives at $x$, or is of bounded variation on an interval $[a,b]$ with $x \in (a,b)$, then the above converges to $(f(x-0)+f(x+0))/2$. You can write the finite integral over $[-R,R]$, interchange integration with the transform integral in order to see why this is true.

So, the answer is "No" to your question. You do not need absolute integrability of $\hat{f}$; the absolutely integrability of $f$ guarantees the local integrability of $\hat{f}$ because it is continuous, and that's all you need under classical conditions where you use the above c.p.v. integral.