For a function $f\in L^1(\mathbb{R})$, its Fourier transform is defined as $$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$$
For a function $f\in L^2(\mathbb{R})$, its Fourier transform is defined as the unique continuous mapping $g:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ that extends the mapping $h:S\rightarrow L^2(\mathbb{R})$, where $S$ is the Schwartz class, and the Fourier transform of a function in the Schwartz class is defined as in the first paragraph. (We may assume that this continuous mapping $g$ exists and is unique.)
Suppose $f\in L^2(\mathbb{R})$, and let $c>0$. Show that $$\lim_{c\rightarrow\infty}\int_{-c}^cf(x)e^{-ixy}dx$$ exists in the $L^2$ sense and is equal to $\hat{f}$ defined above.
Define $f_c(x)$ to be $f(x)$ when $|x|\leq c$ and $0$ when $|x|>c$. Then the limit in question is $$\lim_{c\rightarrow\infty}\int_{-\infty}^\infty f_c(x)e^{-ixy}dx$$
The questions are:
1) Why does this limit exist?
2) Why does it equal $\hat{f}$ defined as the unique extension from the Schwartz class?
We know by the dominated convergence theorem that $\|f_c-f\|_2\rightarrow 0$ as $c\rightarrow\infty$. Might that help?
So $\int_{-c}^c f(x) e^{-ixy} \, dx = \frac1{2i}(H_{-c} - H_c) \hat f(y)$, where $H_c g(x) = e^{icx} Hg (e^{-icx} x)$, and $H$ is the Hilbert transform. The Hilbert transform is known to be an isometry on $L_2$.
To prove what you need, it is sufficient to show $-i H_c g \to g$ in $L_2$ as $c \to -\infty$. But $$ \|g + iH_c g\|_2 = \| 2 \hat g I_{(-\infty,c]} \|_2 .$$
I know I'm invoking some big machinery. But it is not clear to me what you are allowed to assume and not allowed to assume when answering the question.
http://en.wikipedia.org/wiki/Hilbert_transform