1. ## fourier transform/commutation relation

Hello all,

I have been trying to show that:

$(\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)$

I have done the following so far:

$(\digamma D_{c}f)(\gamma)= \int_{ -\infty}^{\infty}(D_{c}f)(x)e^{-2\pi ix\gamma}dx=\int_{-\infty}^{\infty}\frac{1}{\sqrt{c}}f(\frac{x}{c})e^ {-2\pi ix\gamma}dx$

Now if we put $y=\tfrac{x}{c}$ we get:

$\int_{-\infty}^{\infty}\frac{1}{\sqrt{c}}f(y)e^{-2\pi iyc\gamma}cdy=\sqrt{c}\int_{-\infty}^{\infty}f(y)e^{-2\pi iyc\gamma}dy$

How do I continue so that I end up with what I intented to show?

Thanks a million

2. Originally Posted by surjective
I have been trying to show that:

$(\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)$

How do I continue so that I end up with what I intented to show?
What exactly did you intend to show? Rewrite $(D_{\frac{1}{c}}(\mathcal{F}f))(\gamma)$ as an integral, and compare with what you got.

3. ## fourier tranform/commutation relation

Hey,

I wanted to show (intended) that:

$(\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)$

The right side is what I want to arrive at, i.e. I would like to continue from where I left off and arrive at the right-hand-side instead of "going backwards". I seem to be stuck !!?????

4. Originally Posted by surjective
Hey,

I wanted to show (intended) that:

$(\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)$

The right side is what I want to arrive at
I know that; all I wanted you to notice is that you have arrived to the right side. It is just written in a different way. Just like you wrote the left side as an integral, you can do the same for the right side. You'll end up exactly with what your change of variable lead you to. Just one additional "hint" (not that you did it): $D_{\frac{1}{c}}(g)(\gamma)=\sqrt{c}g(\gamma c)$, and take $g(\gamma)=\mathcal{F}f(\gamma)=\int\cdots$.