explain why one can write $\hat{f}(\xi)=\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx$ when $f\in L^2(\mathbb{R})$ Let $f\in L^1(\mathbb{R})$ where the measure is taken to be the Lebesgue measure. The Fourier transform of $f$ is the function $\hat{f}$ defined as 
$$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx \qquad ,\xi\in \mathbb{R}$$

(Plancherel Forumla) If $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ then $\hat{f}\in L^2(\mathbb{R})$ and $\|f\|_2=\|\hat{f}\|_2$

(*) Assuming this result, we can extend the Fourier transform to an isometric operator $L^2(\mathbb{R})\to L^2(\mathbb{R})$

Let $f\in L^2(\mathbb{R})$ be arbitrary. Using the idea (*), explain why one can write $$\hat{f}(\xi)=\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx$$



In this question,


Is the equality $$\lim_{n\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}e^{-i\xi x}f(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx$$ correct?
If the above equality is correct, do we need to show that $\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx&lt;\infty$  i.e.  $\int_{-\infty}^{\infty}|f(x)|dx&lt;\infty$?


How can we solve the question? Thanks!



Update

Let $f_n=\chi_{[-n,n]}f$. Is it enough to show the following steps? 


$f_n\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$
$f_n \to f$ in $L^2$ and $L^1$ as $n\to\infty$
$\hat{f_n}=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n} e^{-i\xi x}f(x)dx$


If the answer yes, my attempt is the following:


Note that $|f_n|&lt;|f|$ implies $\int_{\mathbb{R}}|f_n|^2&lt;\int_{\mathbb{R}}|f|^2&lt;\infty$ . So $f_n\in L^2(\mathbb{R})$. On the other hand we have 
$$\int_{\mathbb{R}}|f_n|=\int_{[-n,n]}|f_n|\leq \sqrt{\int_{[-n,n]}|f_n|^2}\sqrt{\int_{[-n,n]}\textbf{1}^2}=\sqrt{\int_{[-n,n]}|f_n|^2}.\mathcal{L}([-n,n])&lt;\infty$$ So $f_n\in L^1(\mathbb{R})$ as well.
$|f_n-f|^2= \left\{
  \begin{array}{lr}
    0 &amp; \text{on $[-n,n]$}\\
    |f|^2 &amp;  \text{otherwise}
  \end{array}
\right.
$ 


So $$\|f_n-f\|_2^2=\int |f_n-f|^2=\int_{[-n,n]}|f_n-f|^2+\int_{\mathbb{R}-[-n,n]}|f_n-f|^2=\int_{\mathbb{R}-[-n,n]}|f|^2$$
So $\|f_n-f\|_2\to 0$ as $n\to \infty$. Similarly $\|f_n-f\|_1 \to 0$ as $n\to\infty$


By definition, it is obvious.
The following is typically how one would go about equating the $L^2(\mathbb{R})$ extension of the Fourier transform to the classical integral transform.

Theorem: Let $f \in L^2(\mathbb{R})$, and suppose that the following exists for almost every $\xi \in \mathbb{R}$:
$$
           g(\xi) = \lim_{n\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-n}^{n} f(t)e^{-it\xi}dt.
$$
Then $g = \hat{f}$ a.e., where $\hat{f}$ is $L^2(\mathbb{R})$ limit of the integral on the right.

Proof: Suppose $f \in L^2(\mathbb{R})$ and define
$$
             g_n(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}f(t)e^{-i\xi t}dt.
$$
The integral on the right is absolutely convergent for every $n$ because  $\chi_{[-n,n]}f \in L^1$. By Parseval's identity,
$$
          \lim_n \|g_n-\hat{f}\|_{L^2}=0.
$$
By standard results of $L^2$, there is a subsequence $\{ g_{n_k} \}$ that converges pointwise a.e.. to $\hat{f}$. By assumption, $\{ g_n(\xi) \}$ converges pointwise a.e. to $g$. Hence, $g=\hat{f}$ must hold a.e.. $\blacksquare$
$f \in L^2$

$\hat{f}= \lim\limits_{u,v\mapsto \infty}\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$

multiply $f$ by a bump function $g_n$ with compact support in $[-u,v]$

such that $ \lim\limits_{n\mapsto \infty}g_nf=f$ on $[-u,v]$

let $h_n=g_nf$$\quad$ $h_n \in L^1 \bigcap L^2$ and $\hat{h}_n \in L^2 $(Plancherel theorem)

Also $\lim\limits_{n\mapsto \infty}\hat{h}_n=\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$

The good part is that the sequence {$h_n$} is uniformly integrable over $[-u,v]$: this follows from property of bump function.
These imply $\hat{f} \in L^2$

Fourier Inversion theorem:$$
     \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^1(\mathbb{R})} = 0. \;\;\; 
$$

Plancherel theorem:$$
     \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^2(\mathbb{R})} = 0. \;\;\; 
$$

for sufficiently large $n,u,v$:$$\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}_n(s)e^{isx}ds)^2dx \le \delta$$

$$\int_{-\infty}^{\infty}({h}_n-f)^2dx=\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2dx \le \epsilon$$

writing $(a-b)^2=(a-c+c-b)^2=(a-c)^2+(c-b)^2+2(a-c)(c-b)$ and by Cauchy–Schwarz inequality:

$$\int_{-\infty}^{\infty}(f-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds)^2 dx\le \int_{-\infty}^{\infty}((\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds)^2+(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2)dx +(2\delta \epsilon)^{1/2}\le\delta+\epsilon+2(\delta \epsilon)^{1/2}$$