In the book "S. Helgason, The Radon Transform, Birkh¨auser, Boston, Mass., 1980"
You can find that Lemma $2.3$ claims :
$\forall k\in \mathbb{Z}^+: \ $ The integral $$\int_{\mathbb{R}} R(f)(p,w)\ p^k \,dp $$ can be written as a $k^{th}$ degree homogenous polynomial in $w_1,w_2,..., w_n$, here $R(f)$ is the Radon transform of $f\in S(\mathbb{R}^n)$
Without precising what are those $w_1,w_2,..., w_n$ .
So first question: What are those vectors? Does it mean there simply exist some $n$ vectors.
My second question is about the proof, followed in the book; $$\int_{\mathbb{R}} R(f)(p,w)\ p^k \,dp= \int_{\mathbb{R}} \ p^k \int_{\langle x.w\rangle=p}f(x)dm(x) \,dp= \int_{\mathbb{R}^n} f(x)\langle x.w\rangle^kdx$$
I understand this equality but I don't follow why this means that $$\int_{\mathbb{R}^n} f(x)\langle x.w\rangle^kdx= g(w)$$ is a polynomial, and how about homegenous?
And where those $w_1,w_2,..., w_n$ come from ?
The Radon transform $R(f)$ defined here is a function of $w$ and $p$, where $w$ is some unit vector and $p$ is a scalar. For some fixed input $w$, the $w_1, \dots, w_n$ are just the components of this vector in the standard basis for $\mathbb R^n$.
This inner product is then $\left<x,w \right> = x_1w_1 + \dots x_nw_n$ so that $f(x)\left<x,w \right>^k$ is the sum of a bunch of terms of the form $f(x) \phi(x_1,\dots,x_n) \psi(w_1,\dots,w_n)$ where $\phi$ and $\psi$ are both a single $k$th degree term in their respective variables. This integrates "cleanly" w.r.t. $x$ so that you're just left with a bunch of terms that are $k$th degree in the $w_j$'s.