Properties of the Fourier transform

[redtree]

TL;DR Summary

Properties of the Fourier transform of two functions

I was wondering if the following is true and if not, why?

$$
\begin{split}
\hat{f}_1(\vec{k}) \hat{f}_2(\vec{k}) &= \hat{f}_1(\vec{k}) \int_{\mathbb{R}^n} f_2(\vec{x}) e^{-2 \pi i \vec{k} \cdot \vec{x}} d\vec{x}
\\
&= \int_{\mathbb{R}^n} \hat{f}_1(\vec{k}) f_2(\vec{x}) e^{-2 \pi i \vec{k} \cdot \vec{x}} d\vec{x}
\\
&= \mathscr{F}\left[\hat{f}_1(\vec{k}) f_2(\vec{x}) \right]
\end{split}
$$
where
$$
\mathscr{F} \left[ f_n(\vec{x}) \right] = \hat{f}_n(\vec{k})
$$

 

[jbergman]

Your LaTex isn't rendering.

 

[DrClaude]

Your LaTex isn't rendering.

Fixed!

 

[mathman]

It looks like an identity. Am I missing something?

 

[Svein]

The Fourier Convolution Theorem.

 

[redtree]

It’s not the convolution theorem in that only $$\hat{f}_2$$ is Fourier transformed.

I was told by that $$\hat{f}_1$$ cannot be moved into the integral $$\int_{-\infty}^{+\infty} dx$$ and so the equation is not accurate. I disagreed and so posted the question. It seems an identity to me too.

 

[Svein]

The Fourier series transform is a sum over all k, not the value for a single k.

 

[Couchyam]

TL;DR Summary: Properties of the Fourier transform of two functions

I was wondering if the following is true and if not, why?

$$
\begin{split}
\hat{f}_1(\vec{k}) \hat{f}_2(\vec{k}) &= \hat{f}_1(\vec{k}) \int_{\mathbb{R}^n} f_2(\vec{x}) e^{-2 \pi i \vec{k} \cdot \vec{x}} d\vec{x}
\\
&= \int_{\mathbb{R}^n} \hat{f}_1(\vec{k}) f_2(\vec{x}) e^{-2 \pi i \vec{k} \cdot \vec{x}} d\vec{x}
\\
&= \mathscr{F}\left[\hat{f}_1(\vec{k}) f_2(\vec{x}) \right]
\end{split}
$$
where
$$
\mathscr{F} \left[ f_n(\vec{x}) \right] = \hat{f}_n(\vec{k})
$$

From a certain perspective it's only true point-wise in "##\vec k##" space, so it might be misleading. I can't think of any setting off the top of my head where that equation (i.e. ##\mathscr{F}\left[\hat f_1(\vec k) f_2(\vec x)\right] = \hat f_1(\vec k)\hat f_2(\vec k)##) specifically would be useful. The identity ##\mathscr{F}\left[ f\right](\vec k) \equiv \hat f(\vec k)## can be helpful, however, when introducing Fourier analysis to the uninitiated, or improving the flow of a paper/derivation where Fourier analysis is used extensively and intermittently. In general, ##\hat f(\vec k) \hat g(\vec k) = \mathscr{F}\left[f * g\right](\vec k)##, where ##*## is the convolution operator (i.e. ##f * g(x) \equiv \int_y f(y) g(x - y)##, which can be checked with the heuristic "identity" ##\int \frac{dk}{2\pi}e^{ik\cdot x} = \delta(x)##.)