Inverse Fourier transform of an $L^2$ function as limit on balls

Question

$B_m :=\{x \in \Bbb R^n : ||x|| \le m\}$ 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!

Answer 1

Let me reformulate your question: you consider the Fourier multiplier $ L_m=\mathbf 1_{B_m}(D) $ defined by the formula $$ (\mathbf 1_{B_m}(D)f)(x)=\int e^{2iπ x\xi}\mathbf 1_{B_m}(\xi) \hat f(\xi) d\xi. $$ It is obvious that $\lim_{m\rightarrow+\infty}\mathbf 1_{B_m}(D)f=f$ in $L^2(\mathbb R^n)$. In fact since $f$ in $L^2$ is equivalent to $\hat f$ in $L^2$, you can check (for $f\in L^2$), $$ \Vert f-\mathbf 1_{B_m}(D)f\Vert^2_{L^2}=\int_{\vert \xi\vert\ge m}\vert\hat f(\xi) \vert^2 d\xi, $$ which goes to zero, thanks to the Lebesgue Dominated Convergence Theorem. You can also extract a subsequence converging a.e.