Is there any relation between $F(-w)$ and $F(w)$ for the Fourier transform of $f(t)$?

Question

I'm trying to find out how to express $F(-w)$ in terms of $F(w)$ for the Fourier transform of $f(t)$. Thanks!

Answer 1

According to the definition, and for real $\;f\;$ :

$$F(\omega)=\mathscr{F}\{f(x)\}(\omega)=\int_{-\infty}^\infty f(x)e^{-i\omega x}dx=\int_{-\infty}^\infty f(x)\cos\omega xdx-i\int_{-\infty}^\infty f(x)\sin\omega xdx$$

so

$$F(-\omega)=\int_{-\infty}^\infty f(x)e^{-i(-\omega x}dx=\int_{-\infty}^\infty f(x)e^{i\omega x}dx=\int_{-\infty}^\infty f(x)\cos\omega xdx+i\int_{-\infty}^\infty f(x)\sin\omega xdx=$$

$$=\overline{\int_{-\infty}^\infty f(x)\cos\omega xdx-i\int_{-\infty}^\infty f(x)\sin\omega xdx}=\overline{F(\omega)}$$

Answer 2

You can only work out a satisfactory formula when $f$ is real valued. In that case, we have that:

\begin{align} \widehat{f}(-w) &= \int_\mathbb{R^n} f(x)e^{-2\pi i x\cdot (-w)}\, dx \\ &= \int_\mathbb{R^n} f(x)e^{2\pi i x w}\, dx \\ &= \int_\mathbb{R^n} \overline{f(x)e^{-2\pi i x w}}\, dx \\ &= \overline{\int_\mathbb{R^n} f(x)e^{2\pi i x w}\, dx} \\ &= \overline{\widehat{f}(w)}, \end{align}

where the "real-value property" is used to go from line 2 to line 3.

This calculation shows that if $f$ is real-valued, then $\widehat{f}$ has a certain symmetry: an "inversion" in the argument results in a "conjugation" of the output. The same trick works to prove that if $f$ is purely imaginary, then $\widehat{f}(-w) = - \overline{\widehat{f}(w)}$, and one can make out simmilar arguments when one impose some restrictions on the values that $f$ can take.

However, we cannot hope to find any kind of symmetries for Fourier transforms of general complex-valued functions, since the Fourier transform is a bijection from $L^2(\mathbb{R}^n)$ to itself. (In other words, if we apply the Fourier transform to a square-integrable function, the output can be "any" other square-integrable function, which of course could have no symmetry at all!!)

Of course it's possible to find some identities, such as $\widehat{f}(-w) = \overline{\widehat{\overline{f}}(w)}$ (the proof is the same as above), or $\widehat{f}(-w) = \widehat{\widehat{\widehat{f}}}(w)$, but they say nothing in particular about your question.