I don't understand the algebraic steps in the proof of this preposition, I made a $(*)$ symbol where I'm stucked:
Prop: If $T$ is tempered distribution which is homogenous of degree $a$, then its Fourier transform is homogenous of degree $-n-a$
Proof: Let $\phi_{\lambda}(x) = \lambda^{-n}\phi(\lambda^{-1}x)$ then by definition of Fourier transform for $T$ And by using $$(1.18)\qquad \text{if } g(x)=\lambda^{-n}f(\lambda^{-1}x),\text{ then } \hat{g}(\xi)=\hat{f}(\lambda\xi)$$ we have \begin{align*} \hat{T}(\phi_\lambda)&=T(\hat{\phi}(\lambda\cdot))=\lambda^{-n}T(\hat{\phi}_{\lambda^{-1}})=\lambda^{-n-a}T(\hat{\phi})=\lambda^{-n-a}\hat{T}(\phi)\\ \end{align*}
So first what does $\hat{\phi}(\lambda\cdot)$ and $\phi_{\lambda^{-1}}$ mean? Is $\phi_{\lambda^{-1}}=\lambda^{n}\phi(\lambda x)$?
I tried the computational steps: \begin{align*} \hat{T}(\phi_\lambda)&=T(\hat{\phi_\lambda})=T(\hat{\lambda^{-n}\phi(\lambda^{-1}x}))=\lambda^{-n}T(\hat{\phi(\lambda^{-1}x))}\\ \end{align*} and $$\lambda^{-n}T(\phi\hat{}(\lambda^{-1}x))\stackrel{(*)}{=}\lambda^{-n}T(\lambda\hat{\phi})=\lambda^{-n-a}T(\hat{\phi})=\lambda^{-n-a}\hat{T}(\phi)$$
I know that I should apply 1.18 in $(*)$ but what confuses me is that the $\lambda^{-n}$ is outside and that we look at the Fourier transform.
Let's write everything explicitly (I'll use $F[T]$ for Fourier transform).
You have $$\left(F[T],x\to\frac{1}{\lambda^n}\phi\left(\frac x\lambda\right)\right) .$$ By definition of a Fourier transform this is equal to $$\left(T,\xi\to F\left[x\to\frac{1}{\lambda^n}\phi\left(\frac x\lambda\right)\right](\xi)\right)$$ By your formula $1.18$ this is equal to $$\left(T,\xi\to F\left[x\to \phi(x)\right](\lambda \xi)\right).$$ Now we apply the homogenity of $T$ to rewrite the expression above as $$\lambda^{-n-a}\left(T,\xi\to F\left[x\to \phi(x)\right](\xi)\right).$$ Now we put Fourier transform back on $T$: $$\lambda^{-n-a}\left(F[T],x\to \phi(x)\right).$$ In other words, $(F[T],\phi_\lambda(x)) = \lambda^{-n-a}(F[T],\phi(x))$, which gives you the homogenity of $F[T]$.