Originally Posted by

**Deadstar** I suck horribly at using limits and absolute value signs to manipulate integrals but here is my attempt at the attached question. Im pretty close to the answer... I imagine I'm doing something wrong in the absolute value part...

While I was typing this I may have solved it but anyway here goes...

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

b) $\displaystyle \int_{-\infty}^{\infty} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi = $$\displaystyle \int_{|\xi| > 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi + \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi$

Now here's where my 'say it to get the right answer' logic comes in.

Since f is of moderate decrease (or perhaps due to the inequality on $\displaystyle |\hat{f(\xi)}|$...) then we can find a h such that the first integral on the RHS is zero yes?

So assuming the above is true...

We apply absolute values to get...

$\displaystyle f(x+h) - f(x) = \bigg{|} \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi \bigg{|}$

$\displaystyle \leq \int_{|\xi| \leq 1/|h|} |\hat{f(\xi)}| d\xi$ (Can I do that? Take the absolute value inside, thus getting rid of the exponentials..?)

At this point onwards I'm just working things out as I go along...

$\displaystyle \leq \int_{|\xi| \leq 1/|h|} \frac{C}{\xi^{1+\alpha}} d\xi$

$\displaystyle \leq \int \frac{C}{(1/|h|)^{1+\alpha}} d\xi$ (?)

$\displaystyle = \frac{C}{\alpha} |h|^{\alpha} = A|h|^{\alpha}$.

So...

Where do I go wrong (I say that since I have a feeling something is not quite right despite getting the right answer!)