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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?

PJ Miller
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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

Stephen Montgomery-Smith
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  • One more thing. Convergence almost everywhere of your limit is extremely difficult, and is called Carlson's Theorem. – Stephen Montgomery-Smith Nov 17 '13 at 15:09
  • Some more comments. Trying to understand the Fourier transform or Fourier series via partial sums/integrals is a rather difficult way to go. That's why they came up with things like Cesaro sums. In general you need to understand the Fourier transform and how it is an isometry on $L_2$ before engaging on the difficult task of exploring how partial integrals converge. – Stephen Montgomery-Smith Nov 17 '13 at 16:27