I'm very confused with the ideia of extension Fourier transform of $L^1(\mathbb{R}^n)$ to $L^2 (\mathbb{R}^n)$. I start with a $u\in L^1(\mathbb{R}^n)$ and I use the limit and the Banach property to extend the Fourier Transform operator in $L^1(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n),$ it's fine.
In the Plancherel Theorem, I suppose that $u\in L^1(\mathbb{R}^n)\cap L^2 (\mathbb{R}^n)$ to show that
$$\|u\|_{L^2}=\|\widehat{u}\|_{L^2},$$
but I'm ever trapped of hypothesis that $u\in L^1(\mathbb{R}^n)$.
So, can I use the Plancherel Theorem with just $u\in L^2(\mathbb{R}^n)$? How can I do that?
Thanks.
Suppose $f \in L^{2}(\mathbb{R}^{n})$ is $0$ outside a finite cube $[-r,r]^{n}$. Let $\chi_{r}$ be the characteristic function on this cube. Then $f \in L^{1}$ because the following is absolutely integrable $$ 2|f|\chi_{r} \le |f|^{2}+\chi_{r}^{2}. $$ Therefore $f_{r} = f\chi_{r}$ is in $L^{1}$ and has a classical Fourier transform. As $r\rightarrow\infty$, the function $f_{r}$ converges in $L^{2}$ to $f$; hence by Plancherel's Theorem, $\widehat{f_{r}}$ converges in $L^{2}$ to a unique function that one defines to be $\hat{f}$. The limit can be taken in any way that the trunction converges in $L^{2}$ to the original function, and the result is the same $\hat{f}$. But, this is a reasonable way to define it: $$ \hat{f}(s_1,s_2,\cdots,s_n)=\lim_{r\rightarrow\infty}\frac{1}{(2\pi)^{n/2}}\int_{-r}^{r}\int_{-r}^{r}\cdots\int_{-r}^{r} f(x_1,x_2,\cdots,x_n)e^{-is_1x_1}e^{-is_2x_2}\cdots e^{-is_nx_n} $$ The convergence is in $L^{2}$. You can do this in different ways, but you end up with the same $L^{2}$ function. This definition does not require any mention of anything other than $f \in L^{2}$.