n} +
\inf_{\theta \in S^{n-1}} g(t \theta) \ \ \ \text{for all} \ \  t
\geq 0. \label{eq_750}$$ \[lem\_fourier\]

*Proof:* We will show that the Fourier transform of $f$ is almost spherically-symmetric. As usual, we define $$\hat{f}(\xi) = \int_{\RR^n} e^{-2 \pi i \langle \xi, x \rangle}
f(x) dx \ \ \ \ (\xi \in \RR^n),$$ where $i^2 = -1$. Let $r > 0$, and fix $\xi_1, \xi_2 \in \RR^n$ with $|\xi_1| = |\xi_2| = r$. Denote by $E_1 = \RR \xi_1, E_2 = \RR \xi_2$ the one-dimensional subspaces spanned by $\xi_1, \xi_2$, respectively. From (\[eq\_621\_\]) of Section \[section2\] we see that $\pi_{E_j}(f)(t \xi_j/|\xi_j|) = \frac{\partial}{\partial t} M_{f}
(\xi_j / |\xi_j|, t)$ for $j=1,2$ and for all $t$ in the interior of the support of the log-concave function $t \mapsto
\pi_{E_j}(f)(t \xi_j/|\xi_j|)$. By integrating by parts we obtain $$\begin{aligned}
\nonumber \label{f_xi} \lefteqn{\hat{f}(\xi_1) - \hat{f}(\xi_2)=
\int_{-\infty}^{\infty} \left[ \pi_{E_1}(f) \left(t
\frac{\xi_1}{|\xi_1|} \right) - \pi_{E_2}(f) \left(t
\frac{\xi_2}{|\xi_2|} \right) \right] e^{-2 \pi i r t} dt } \\ & =
& 2 \pi i r \int_{-\infty}^{\infty} \left[M_{f} \left(
\frac{\xi_1}{|\xi_1|}, t \right) - M_{f} \left(
\frac{\xi_2}{|\xi_2|}, t \right)  \right] e^{-2 \pi i r t} dt,
\phantom{aaaaaaaa}\end{aligned}$$ as the boundary terms clearly vanish. From Lemma \[lem\_1028\] we 