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Fourier Transform of $\mathcal{L_1}$-functions is defined as follows:

$\hat{f}(\omega)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)\cdot e^{i\omega x} dx $.

I can think of three ways to define an Improper Integral of the form $\int _{-\infty}^{\infty}$:

  1. Choose arbitrary $x\in \mathbb R$ and split $\int_{-\infty}^{x} + \int_{x}^{\infty} $

  2. $\lim _{n\to \infty} \int_{-n}^{n}$

  3. $\lim _{n\to \infty, m\to -\infty } \int _{m}^{n}$

Outside the context of fourier analysis, i'm using definition 3 for improper integrals of this form. 1 is freely used in calculations of fourier transforms, and 2 is used in the fourier inversion theorem (At least at some of it's formulations).

How is the integral in fourier transform defined? Perhaps the methods above are equivalent for $\mathcal L_1$-functions?

Thanks in advance!

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I would simply define the Fourier transform $\int_{-\infty}^\infty f(x) e^{i \omega x}$ to be the Lebesgue integral of $f(x) e^{i\omega x}$ over the whole of $\mathbb R$. Since $|f(x) e^{i\omega x}| = |f(x)|$, and since $f(x)$ is in $\mathcal L_1$ by assumption, we can be sure that this integral is well-defined.

My suggestion is equivalent to all of your suggestions (1), (2) and (3).

  • Since $f(x) e^{i \omega x} = f(x) e^{i \omega x} \chi_{(-\infty, a]} + f(x) e^{i \omega x} \chi_{[a, \infty)}$ for any fixed $a \in \mathbb R$ (where $\chi_A$ always denotes the indicator function on $A \subset \mathbb R$), we immediately get $$ \int_{-\infty}^\infty f(x) e^{i \omega x} = \int_{-\infty}^a f(x) e^{i \omega x} + \int_{a}^\infty f(x) e^{i \omega x}.$$ So my suggestion is the same as your suggestion number (1).

  • Next, observe that $f(x)e^{i\omega x} \chi_{(-n,n)}$ is dominated by $|f(x)|$, which is integrable by assumption. By the dominated convergence theorem, we learn that $$\int_{-\infty}^\infty f(x) e^{i \omega x} = \lim_{n \to \infty} \int_{-n}^n f(x) e^{i \omega x}$$ So my suggestion is the same as your suggestion number (2).

  • We can also apply the dominated convergence theorem to the sequence of functions $f(x) e^{i\omega x} \chi_{(-m,n)}$ with $n$ fixed and $m$ varying in $ \mathbb N$ to show that $$\lim_{m \to - \infty} \int_{m}^n f(x) e^{i\omega x} = \int_{-\infty}^n f(x) e^{i \omega x}.$$ Applying dominated convergence a second time, using the sequence of functions $f(x) e^{i\omega x} \chi_{(-\infty,n)}$, with $n \in \mathbb N$, we learn that $$\lim_{n \to \infty} \left( \lim_{m \to - \infty} \int_{m}^n f(x) e^{i\omega x} \right) = \lim_{n \to \infty} \int_{-\infty}^n f(x) e^{i \omega x} = \int_{-\infty}^{\infty} f(x) e^{i \omega x}.$$ So my suggestion is the same as your suggestion number (3).

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  • $\begingroup$ Thank you for the clear and imformative answer :) $\endgroup$ – user333618 Apr 4 '17 at 15:45

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