Is $f(x)=\lim_{R\rightarrow \infty}\int_{-R}^{R}\hat{f}(w)e^{2i\pi xw}dw$?

Question

Let $f$ be a function satisfying $$\int_{-\infty}^{\infty}|f(x)|dx<\infty.$$

Is it true that for almost every $x\in\mathbb{R}$, $$f(x)=\lim_{R\rightarrow+\infty}\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw?$$

Here $\hat{f}$ denotes the Fourier transform of $f$.

Thanks for any comment or any suggestions.

Answer 1

Since the FT of $f(x)$ as $\hat f(\omega)$ exists, we can write:$$\hat f(\omega)=\int_{-\infty}^\infty f(x)e^{-i2\pi\omega x}dx$$and by substituting we obtain$$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw{=\int_{-R}^{R}\int_{-\infty}^\infty f(x_1)e^{-i2\pi\omega x_1}e^{2i\pi x w}dx_1dw\\=\int_{-\infty}^\infty f(x_1)\int_{-R}^{R} e^{-i2\pi\omega x_1}e^{2i\pi x w}dwdx_1\\= \int_{-\infty}^\infty f(x_1){\sin 2\pi R(x-x_1)\over R(x-x_1)}dx_1 }$$since the latter integral is absolutely convergent we can write: $$\lim_{R\rightarrow+\infty}\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw=\int_{-\infty}^\infty f(x_1)\lim_{R\rightarrow+\infty}{\sin 2\pi R(x-x_1)\over R(x-x_1)}dx_1\\=\int_{-\infty}^\infty f(x_1)\delta(x-x_1)dx_1=f(x)$$where $\delta(x)$ denotes the Dirac delta function.

Answer 2

A simpler case is when $f$ is $L^1$ and locally $a$-Hölder continuous, from $$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw = \int_{-\infty}^\infty f(x-y)\frac{\sin(2 \pi R y)}{\pi y}dy$$

we obtain that it converges to $f$ locally uniformly.

Note that $\int_{|y|> b}|\frac{f(x-y)}{\pi y} |dy< \infty$ implies $\lim_{R \to \infty}\int_{|y|> b}\frac{f(x-y)}{\pi y} \sin(2\pi R y)dy = 0$ thus we care only of the behavior on a neighborhood of $x$.