Deriving Fourier inversion formula from Fourier series

Let $g\in C_0^{\infty}(\mathbb{R})$ (infinitely differentiable with compact support), and let $$\hat{g}(y)=\int_{-\infty}^\infty g(x)e^{-ixy}dx$$ Assume that $\hat{g}$ is in the Schwartz class. Prove that $$g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{ixy}dy$$

We may use the result that if $f\in C^{\infty}(\mathbb{R})$ is a periodic function of period $2L$, then $$\hat{f}(x)=\sum_{n=-\infty}^\infty \left(\dfrac{1}{2L}\int_{-L}^Lf(y)e^{-in\pi y/L}dy\right)e^{i\pi nx/L}$$

I'm trying to follow Steven Stadnicki's hint. Since $g$ has compact support, let $N$ be such that $g(x)=0$ for all $|x|>N$. Choose $L>N$, and let $f_L(x)=g(x)$ for $|x|\leq N$ and extend $f_L(x)$ periodically with period $2L$ to all of $\mathbb{R}$.

Then we have $$\hat{f_L}(x)=\sum_{n=-\infty}^\infty \left(\dfrac{1}{2L}\int_{-L}^Lf(y)e^{-in\pi y/L}dy\right)e^{i\pi nx/L}$$

If I send $L$ to $\infty$, in some sense I get the function $g$. But I'm still confused how the Fourier coefficients of $f_L$ will translate to the coefficients of $g$.

Solution 1:

My outline of solution:

Replacing $\hat{g}(y)$ by the integral, we will obtain $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{ixy}dy= \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(z)e^{iy(x-z)}dz\right)dy$$

Next we multiply a term $e^{-\frac{\epsilon^2y^2}{4}}$ into the integration, and consider the limit as $\epsilon\to0$,

$$A(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(z)e^{-\frac{\epsilon^2y^2}{4}}e^{iy(x-z)}dz\right)dy$$ After simplifying, we get $$A(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}dy$$ Since $|\hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}|\leq|\hat{g}(y)|$, $$\lim_{\epsilon\to0}A(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\lim_{\epsilon\to0} \hat{g}(y)e^{-\frac{\epsilon^2y^2}{4}}e^{ixy}dy=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(y)e^{ixy}dy$$...