Estimate involving almost orthogonality Let $\{x_k\}$ be a $\frac{1}{R}$-separated set of points on $S^{d-1}$. Then
$$
\left\|\sum_ka_ke^{ix_k\cdot\xi}\right\|_{L^2\left(B\left(0,R\right)\right)}\lesssim R^{d/2}
\left(\sum_k|a_k|^2\right)^{1/2},
$$
where $B\left(0,R\right)$ is the ball of radius $R$ in $\mathbb{R}^d$. 
The author claims that the result follows from the almost orthogonality of $\{e^{ix_k\cdot\xi}|_{B\left(0,R\right)}\}$. I tried expanding the square, but I had troubles estimating mixed products. 
I tried also using Cotlar-Stein lemma, that is, define the operators $T_k:l^{N}\to L^2\left(B\left(0,R\right)\right)$ as $T_k\left(a\right)=a_ke^{ix_k\cdot\xi}$, where $N$ is the number of points on $S^{d-1}$. Then $\|T_kT_l^*\|=c_d\delta_{kl}R^{d}$, where $c_d$ is the area of the unit ball, and $\|T_k^*T_l\|=R^d|\int_{B\left(0,1\right)}e^{iR\left(x_k-x_l\right)\cdot\xi}\,d\xi|\lesssim R^{\left(d-1\right)/2}\frac{1}{|x_k-x_l|^{\left(d+1\right)/2}}$ if $k\neq l$ and  $\|T_k^*T_k\|\sim R^d$. 
We have $\sum_l\|T_kT_l^*\|^{1/2}\sim R^{d/2}$, but $\sum_l\|T_k^*T_l\|^{1/2}\lesssim R^{\left(d-1\right)/4}\left(R^{\left(d+1/4\right)}+\sum_{\substack{l=1\\ l\neq k}}^N\frac{1}{|x_k-x_l|^{\left(d+1\right)/4}}\right)$ and in the worst case $N\sim R^{d-1}$, so the summation term could be rather large.
It seems that the author proves this lemma easily, but I don't know how. Thanks! 
 A: The following proof of your claim uses a somewhat different form of
"orthogonality", namely of translations of a certain function
on the Fourier side.
In the following, I use the version $\widehat{f}\left(\xi\right)=\int f\left(x\right)e^{-2\pi i\left\langle x,\xi\right\rangle }\, dx$
of the Fourier transform. It might be that I miss some factors of
$2\pi$ below. I hope that can be excused.
Let us fix $\varphi\in\mathcal{S}(\mathbb{R}^{d})$ with
the property that $\widehat{\varphi}\in C_{c}^{\infty}\left(B_{1}\left(0\right)\right)$
with $\widehat{\varphi}\geq0$ and $\varphi\left(0\right)=\int\widehat{\varphi}\, dx=2$.
Here, the first equality comes from Fourier inversion. Existence of
such a function is a standard result.
As $\varphi$ is continuous with $\varphi\left(0\right)=2$, there
is some $r>0$ such that $\left|\varphi\left(x\right)\right|\geq1$
holds for all $\left|x\right|\leq r$. By shrinking $r$, we can assume
$4\pi r\leq1$. Then, the family $\left(x_{k}\right)_{k}$ is a fortiori
also $\frac{4\pi r}{R}$-separated.
Consider $$\varphi_{R}\left(x\right):=\left(\frac{r}{R}\right)^{d}\cdot\varphi\left(\frac{r}{R}x\right).$$
We then have (easy calculation) $\left\Vert \varphi_{R}\right\Vert _{2}=\left(\frac{r}{R}\right)^{d/2}\cdot\left\Vert \varphi\right\Vert _{2}$,
as well as $\left|\varphi_{R}\left(x\right)\right|\geq\left(\frac{r}{R}\right)^{d}$
for $\left|x\right|\leq R$ and the Fourier transform of $\varphi_{R}$
is given by
$$
\widehat{\varphi_{R}}\left(\xi\right)=\widehat{\varphi}\left(\frac{R}{r}x\right)
$$
which is supported in $B_{r/R}\left(0\right)$.
For $\xi\in\mathbb{R}^{d}$ and $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$,
write $\left(M_{\xi}f\right)\left(y\right):=e^{2\pi i\left\langle \xi,y\right\rangle }$
and $\left(L_{\xi}f\right)\left(y\right):=f\left(y-\xi\right)$. As
modulation corresponds to translation on the Fourier side (this is
an easy calculation), we have, for $k\in\left\{ 1,\dots,N\right\} $:
$$
\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}=L_{\frac{x_{k}}{2\pi}}\widehat{\varphi_{R}}.
$$
As the family $\left(x_{k}\right)_{k}$ is $\frac{4\pi r}{R}$-separated,
we get
\begin{eqnarray*}
{\rm supp}\left(\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right)\cap{\rm supp}\left(\widehat{M_{\frac{x_{n}}{2\pi}}\varphi_{R}}\right) & = & {\rm supp}\left(L_{\frac{x_{k}}{2\pi}}\widehat{\varphi_{R}}\right)\cap{\rm supp}\left(L_{\frac{x_{n}}{2\pi}}\widehat{\varphi_{R}}\right)\\
 & \subset & \left[B_{r/R}\left(0\right)+\frac{x_{k}}{2\pi}\right]\cap\left[B_{r/R}\left(0\right)+\frac{x_{n}}{2\pi}\right]\\
 & = & \frac{1}{2\pi}\cdot\left(\left[B_{2\pi r/R}\left(0\right)+x_{k}\right]\cap\left[B_{2\pi r/R}\left(0\right)+x_{n}\right]\right)=\emptyset.
\end{eqnarray*}
Finally, for $f_{1},\dots,f_{n}\in L^{2}(\mathbb{R}^{d})$ with
disjoint supports, we have
$$
\left\Vert \sum f_{j}\right\Vert _{2}^{2}=\int\left|\sum f_{j}\right|^{2}\, dx=\int\sum\left|f_{j}\right|^{2}\, dx=\sum\left\Vert f_{j}\right\Vert _{2}^{2}.\qquad\left(\dagger\right)
$$
We now employ $\left(\dagger\right)$, the Plancherel theorem, as
well as $\left|\varphi_{R}\left(x\right)\right|\geq\left(\frac{r}{R}\right)^{d}$
on $B_{R}\left(0\right)$, to derive
\begin{eqnarray*}
\left\Vert \sum_{k}a_{k}e^{i\left\langle x_{k},\xi\right\rangle }\right\Vert _{L^{2}\left(B_{R}\left(0\right)\right)} & \leq & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}e^{i\left\langle x_{k},\xi\right\rangle }\varphi_{R}\left(\xi\right)\right\Vert _{L^{2}}\\
 & = & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}\cdot M_{\frac{x_{k}}{2\pi}}\varphi_{R}\right\Vert _{L^{2}}\\
 & \overset{\text{Plancherel}}{=} & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}\cdot\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right\Vert _{L^{2}}\\
 & \overset{\left(\dagger\right)}{=} & \left(\frac{R}{r}\right)^{d}\cdot\sqrt{\sum_{k}\left\Vert a_{k}\cdot\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right\Vert _{L^{2}}^{2}}\\
 & \overset{\text{Plancherel}}{=} & \left(\frac{R}{r}\right)^{d}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}\cdot\left\Vert \varphi_{R}\right\Vert _{2}^{2}}\\
 & = & \left(\frac{R}{r}\right)^{d}\cdot\left(\frac{r}{R}\right)^{d/2}\cdot\left\Vert \varphi\right\Vert _{2}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}}\\
 & = & \frac{\left\Vert \varphi\right\Vert _{2}}{r^{d/2}}\cdot R^{d/2}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}}.
\end{eqnarray*}
This proves the claim, as long as we allow the implied constant to
depend on the dimension $d$, as $\varphi,r$ can be chosen once and
for all for each dimension $d\in\mathbb{N}$. Also note that we did
not make use of the assumption $\left|x_{k}\right|=1$ for all $k$.
