- #1

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I should prove:

[tex] \delta'(\lambda x) = \dfrac{1}{\lambda \vert \lambda \vert} \delta(x), [/tex]

where lambda is just a constant.

If we make use of the scaling property and the definition of the distributional derivative, we find:

[tex] \left( \delta'(\lambda x), f \right) = \dfrac{1}{\vert \lambda \vert} \left( \delta', f(x/\lambda) \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} \left( \delta, f'(x/\lambda) \right) = \dfrac{1}{\lambda \vert \lambda \vert} \left( \delta', f(x/\lambda) \right) [/tex]

Because this is true for all testfunctions f, we have shown the identity.

Now I think that my last step is wrong.

Because of

[tex] f'(x/\lambda) = \left( \dfrac{\partial f}{\partial x} \right) (x/\lambda) \neq \dfrac{\partial}{\partial x} \left( f(x/\lambda) \right) = \dfrac{1}{\lambda} f'(x/\lambda) [/tex]

there should be an additional factor of lambda if I 'shift' the derivative back to the delta distribution.

You can also see this by (formally) writing the last step:

[tex] \dfrac{(-1)}{\lambda \vert \lambda \vert} \left( \delta, f'(x/\lambda) \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} \int_{-\infty}^{\infty} \delta(x) f'(x/\lambda) \, \mathrm{d}x = \dfrac{1}{\vert \lambda \vert} \int_{-\infty}^{\infty} \delta'(x) f(x/\lambda) \, \mathrm{d} x = \dfrac{1}{\vert \lambda \vert} \left( \delta', f(x/\lambda) \right), [/tex]

where I used integration by parts.

So something is really wrong here.

Maybe someone could help me.