1. ## Fourier Inversion

Given that f is a function of moderate decrease, $\displaystyle \hat{f}(\xi)$ is continuous and also satisfies,

$\displaystyle |\hat{f}(\xi)| \leq \frac{C}{|\xi|^{1+\alpha}}$.

Use the inversion theorem to derive an expression for $\displaystyle f(x+h) - f(x)$. It should be in the form of an integral.

How would I do this? Would I use a change of variable?
Just typed out long possible solution but then realized inversion theorem is $\displaystyle d \xi$ not $\displaystyle dx$...

Given that f is a function of moderate decrease, $\displaystyle \hat{f}(\xi)$ is continuous and also satisfies,

$\displaystyle |\hat{f}(\xi)| \leq \frac{C}{|\xi|^{1+\alpha}}$.

Use the inversion theorem to derive an expression for $\displaystyle f(x+h) - f(x)$. It should be in the form of an integral.

How would I do this? Would I use a change of variable?
Just typed out long possible solution but then realized inversion theorem is $\displaystyle d \xi$ not $\displaystyle dx$...
Well, by the inversion formula, $\displaystyle f(x+h)-f(x)=\frac{1}{2\pi}\int (e^{-i\xi(x+h)}-e^{-i\xi x})\hat{f}(\xi)d\xi$ . You can factor by $\displaystyle e^{-i\xi x}$, but that's almost all you can do.

3. Yeah that's what I had. Thought there might be more to it, oh well!

4. Actually that's not what I got I never read it right!

I think I'm getting majorly confused here. We have the Fourier Inversion defined as;

$\displaystyle f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{2 \pi i x \xi} d \xi$...

So wouldn't it be.

$\displaystyle f(x+h) - f(x) = \int_{-\infty}^{\infty} (e^{2 \pi i (x+h) \xi} - e^{2 \pi i x \xi}) \hat{f}(\xi) d \xi$.

Where does the $\displaystyle \frac{1}{2 \pi}$ come from?

Where does the $\displaystyle \frac{1}{2 \pi}$ come from?
There are at least three commonly used definitions of the Fourier transform, and I simply used another one... (namely $\displaystyle \mathcal{F}(f)(\xi)=\int e^{it\xi}f(t)dt$, without $\displaystyle 2\pi$ in the exponent) You switch from one to another by a change of variable. You formula is correct with you definition.