Let $F$ a tempered distribution, in my textbook the transform of $F$ is defined by $\hat{f}[ \phi]=F[\hat{\phi}]$ , having said that l'm interested in prove the following property of the Fourier transform:
$\mathbf{F}(F(tx))=|t|^{-n} \hat{f}(t^{-1} \xi)$
My attempt:
$(\mathbf{F}[F(tx)])[\phi]=F[\hat{(t\phi)}]=F[|t|^{-n} \hat{\phi }(t^{-1} \xi)]=|t|^{-n}F[\hat{\phi} (t^{-1} \xi)]=|t|^{-n} \hat{f}(t^{-1} \xi )$
This is correct? I am not sure that I have applied the definitions correctly. Best regards.
Added: n is the number of dimensions
How I would write it
I here use the notation $\langle u, \varphi \rangle$ for the application of the distribution $u$ on the test function $\varphi.$
Introduce the operator $\lambda_t$ defined by $(\lambda_t f)(x) = f(tx).$ For a distribution $u$ we defined $\lambda_t u$ by $\langle \lambda_t u, \varphi \rangle = |t|^{-n} \langle u, \lambda_{1/t} \varphi \rangle.$
For a test function we have $\lambda_t\hat\varphi = |t|^{-n}\widehat{\lambda_{1/t}\varphi}.$ Using the above formulas, for a distribution we get $$ \langle \widehat{\lambda_t u}, \varphi \rangle = \langle \lambda_t u, \hat\varphi \rangle = |t|^{-n} \langle u, \lambda_{1/t}\hat\varphi \rangle = |t|^{-n} \langle u, |1/t|^{-n} \widehat{\lambda_t\varphi} \rangle \\ = \langle u, \widehat{\lambda_t\varphi} \rangle = \langle \hat u, \lambda_t\varphi \rangle = |t|^{-n} \langle \lambda_{1/t}\hat u, \varphi \rangle , $$ i.e. $\widehat{\lambda_t u} = |t|^{-n} \lambda_{1/t}\hat u.$