$B_m :=\{x \in \Bbb R^n : ||x|| \le m\}$ and $\mathscr{F}f$ denotes the $L^2$ fourier transform of an $f \in L^2(\Bbb R^n)$.

I am trying to show that

If $f \in L^2(\Bbb R^n)$ then $f(y)=\lim_{m \to \infty} \int_{B_m} \mathscr{F}f(x)e^{2\pi ix.y} dx$

$f \in L^2(\Bbb R^n) \implies \mathscr{F}f \in L^2(\Bbb R^n)$

Now define, $g_m(x)=\mathscr{F}f(x) \chi_{B_m}(x)$

Then $g_m \in L^1(\Bbb R^n) \cap L^2(\Bbb R^n)$ and $g_m \to \mathscr{F}f$ pointwise.

Since, $g_m \in L^1(\Bbb R^n)$ take their Fourier inverse because for them I have the expression for Fourier inversion which should be handy.

So If I could show that the inverse Fourier transforms of $g_m$ converge to our $f$ pointwise, we are through! How to prove this?

I also tried by Schwarz density kind of arguments but I am getting double sequences and too hard to handle!

$B_m :=\{x \in \Bbb R^n : ||x|| \le m\}$ and $\mathscr{F}f$ denotes the $L^2$ fourier transform of an $f \in L^2(\Bbb R^n)$.

I am trying to show that

If $f \in L^2(\Bbb R^n)$ then $f(y)=\lim\limits_{m \to \infty} \int_{B_m} \mathscr{F}f(x)e^{2\pi ix.y} dx$ a.e.

$f \in L^2(\Bbb R^n) \implies \mathscr{F}f \in L^2(\Bbb R^n)$

Now define, $g_m(x)=\mathscr{F}f(x) \chi_{B_m}(x)$

Then $g_m \in L^1(\Bbb R^n) \cap L^2(\Bbb R^n)$ and $g_m \to \mathscr{F}f$ pointwise a.e.

Since, $g_m \in L^1(\Bbb R^n)$ take their Fourier inverse because for them I have the expression for Fourier inversion which should be handy.

So If I could show that the inverse Fourier transforms of $g_m$ converge to our $f$ pointwise a.e., we are through! How to prove this?

I also tried by Schwarz density kind of arguments but I am getting double sequences and too hard to handle!