I am studying Mattila's book "Fourier Analysis and Hausdorff Dimension" and I do not understand how to reach the first inequality in the proof of Theorem 4.2. It is the following:
Let $2<s<\text{dim }A$ and choose $\mu$ with compact support contained in $A$ such that $0<\mu(A)<\infty$ and $I_s(\mu)<\infty$, where $I_s(\mu)$ is the energy integral of $\mu$. We know this measure exists due to what was studied in chapter 2.
Let $P_e: \mathbb{R}^n\to \mathbb{R}$ be the projection $x\mapsto e\cdot x$ for $e\in S^{n-1}$ and define the measure $\mu_e$ in $\mathbb{R}$ as $\mu_e(B) = \mu(P_e^{-1}(B))$ for $B\subset\mathbb{R}$. With all this we have
$$ \int_{S^{n-1}}\int_{\mathbb{R}} |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e \leq 2\int_{S^{n-1}}\int_1^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e + 2\mu(\mathbb{R^n})\sigma^{n-1}(S^{n-1}). $$
What I don't get is how to obtain the second term, I do understand all that follows and why it is necessary to "cut" the integral at $1$, but I cannot get the $2\mu(\mathbb{R^n})\sigma^{n-1}(S^{n-1})$.
$$ \int_{S^{n-1}}\int_{\mathbb{R}} |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e = \int_{S^{n-1}}\int_{1}^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e + \int_{S^{n-1}}\int_{-\infty}^1 |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e, $$
so to get the $2\mu(\mathbb{R^n})\sigma^{n-1}(S^{n-1})$ I need to bound the second term, which is:
$$ \int_{S^{n-1}}\int_{-\infty}^1 |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e = \int_{S^{n-1}}\int_{-\infty}^1 |\widehat{\mu}(re)|\,dr\,d\sigma^{n-1}e = \int_{S^{n-1}}\int_{-1}^\infty |\widehat{\mu}(-te)|\,dt\,d\sigma^{n-1}. $$
Edit:
Using MathWonk's suggestion I get $$ \int_{S^{n-1}}\int_{-\infty}^1 |\widehat{\mu}(re)|\,dr\,d\sigma^{n-1}e \leq \int_{S^{n-1}}\int_{-\infty}^1|\widehat{\mu}(0)|\,dr\,d\sigma^{n-1}e\leq \sigma^{n-1}(S^{n-1})\mu(\mathbb{R}^n)\int_{-\infty}^1dr, $$ which is not bounded.
I have been stuck with this inequality for about a week now, any help is greatly appreciated!!
I was able to solve it using MathWonk's suggestion that $\widehat{\mu}(0)\geq \widehat{\mu}(\xi)$ for each $\xi$.
The idea was not to separate the integral between $(-\infty,1)$ and $(1,\infty)$, but to do the following:
\begin{align*} \int_{S^{n-1}}\int_{-\infty}^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e &= 2\int_{S^{n-1}} \int_0^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e\\ &= 2\int_{S^{n-1}} \int_1^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e + 2\int_{S^{n-1}}\int_0^1 |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e\\ &\leq 2\int_{S^{n-1}} \int_1^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e + 2\int_{S^{n-1}}\int_0^1 |\widehat{\mu_e}(0)|\,dr\,d\sigma^{n-1}e\\ &= 2\int_{S^{n-1}} \int_1^\infty |\widehat{\mu_e}(r)|\,dr\,d\sigma^{n-1}e + 2\sigma^{n-1}(S^{n-1})\mu(\mathbb{R}^n). \end{align*}