In the context of Fourier transforms, is the following true?
$$ \mathcal{F} \left( F(t) \right) =2\pi \, f(-\omega ) $$
If it is not, what small changes can I make to the equation to make it true?
I found the problem like that, but I think both uppercase $F$'s are supposed to denote a Fourier transform. The closest thing I have found so far is
this section in
Wikipedia.
There, I found this:
$$ f(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbb {R} ^{n}}{\hat {f}}(\omega
)e^{i\omega \cdot x}\,d\omega $$
However, I am not sure what restrictions I need to make to use that last equation, or whether it is useful at all. I do not even think I completely understand it.
So, how can I make the equation at the top true?
I think this is what you're asking:
Let $f$ be a nice function (say a Schwartz function). If we define the Fourier transform as $$\mathcal{F} f(\xi)=\int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}\, dx,$$ then we have the Fourier inversion formula: $$\mathcal{F}^{-1}=\frac{1}{(2\pi)^n} R\mathcal{F},$$ where $R$ is an operator defined by $Rf(x)=f(-x).$ In particular, we get the formula you wrote in your post: $$f(x)=\frac{1}{(2\pi)^n} \int\limits_{\mathbb{R}^n} \mathcal{F}f(\xi)e^{ix\cdot\xi}\, d\xi.$$
For the formula you look like you want, you can get something like $$(2\pi)^n\mathcal{F}^{-1} f(x)=\mathcal{F}f(-x).$$
As for when this inversion holds, you can get pretty sharp results, depending on the generality you want to work in.